# ASF Spoilers Thread



## cynic (20 May 2015)

Ves said:


> Love this sort of thing.  Haven't posted any responses,  because usually when I've had the chance to check someone else has long figured it out.
> 
> Just a thought.... not sure if it's a problem for others,  but it's hard to open this thread and not be exposed to spoilers  (ie.   someone else's thoughts or even the answer).
> 
> Would it be possible to have two separate threads - one exclusively for the logic puzzles / riddles  and another for discussions of the processes and eventual answers? Shouldn't be too hard to cross-reference between the two with the right headings or post references  (Tech/A used to do it in his charting threads).   I hope that something like this would keep it in the spirit of community collaboration,  because it's great to see,  especially in the often maligned general chat section of ASF.






skyQuake said:


> Correct!
> 
> Well done ASF, you have collectively passed the recommend requirement of 9 and over
> 
> ...






cynic said:


> I think I know the answer to #5, but I'll wait for creation of that other thread.
> 
> Edit: and #6 also.
> 
> 2nd edit: and #5 as well.



My  answers:
Answer to #6 the fridge magnet can be used to test if the purport replica is made of iron.
Answer to #5 the fingerprints will be on the bottle of cleaner - he/she couldn't spray it with itself and couldn't use it without leaving his/her fingerprints.
Answer to #4 The woman in pink. All other patrons are dressed in black.


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## keithj (20 May 2015)

#4 The waiter. When the meal arrives, the waiter is serving oysters, but claiming they are escargots.


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## cynic (20 May 2015)

keithj said:


> #4 The waiter. When the meal arrives, the waiter is serving oysters, but claiming they are escargots.



Darn it! 
Foiled again, and all because I don't happen to speak French!


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## skc (20 May 2015)

keithj said:


> #4 The waiter. When the meal arrives, the waiter is serving oysters, but claiming they are escargots.




You are probably correct. Except the restaurant is serving a special creation by a master chef today... Escargots stewed in oyster shells with citrus dressing.


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## skyQuake (20 May 2015)

4,5,6 correct! On to the next batch


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## galumay (20 May 2015)

Which #4 is correct? I cant see why the waiter is correct - just because he calls the oysters, snails. Likewise if its the girl in pink, what is the logic?


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## skyQuake (20 May 2015)

galumay said:


> Which #4 is correct? I cant see why the waiter is correct - just because he calls the oysters, snails. Likewise if its the girl in pink, what is the logic?




Those clams/oysters should not be escargots - the waiter is obviously a phony


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## cynic (20 May 2015)

#4 Escargot is a French dish of cooked snails.

Answer to # 12 : Use the wrench to remove car tyre tubes and tie them to the bag of coins with the rope.


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## keithj (20 May 2015)

galumay said:


> Which #4 is correct? I cant see why the waiter is correct - just because he calls the oysters, snails.



A real waiter would not make such an elementary mistake, therefore he cannot be a real waiter, and can only be bank robber.  Gotta think like a 9yo


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## keithj (20 May 2015)

#7 make a snowball containing the $$ & throw it back over ?


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## keithj (20 May 2015)

#10.  He's not wearing bowling shoes.... everyone else is.


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## keithj (20 May 2015)

cynic said:


> Answer to # 12 : Use the wrench to remove car tyre tubes and tie them to the bag of coins with the rope.




But not necessarily in that order ?


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## galumay (20 May 2015)

skyQuake said:


> Those clams/oysters should not be escargots - the waiter is obviously a phony




Yep, i was trying to overcomplicate it!


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## galumay (20 May 2015)

#8 the alarm clock in the bookshelf

#11 elephant has a bandage on its trunk!

#9 has me stuck!


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## keithj (20 May 2015)

#11.  It's a bit tenuous but....

The only one actually in pain is the croc.  The rest are there for routine appointments.


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## galumay (20 May 2015)

bellenuit said:


> A famous classic puzzle.
> 
> *You are a criminal found guilty of a serious crime and have been sentenced to death. The judge, being a puzzle freak, decides to give you a chance of avoiding death.
> 
> ...




The implication is that in opening the remaining box that he knows for sure doesnt have the pardon, the other remaining box *does* contain the pardon, so switching boxes would appear to increase the chance of freedom!


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## keithj (20 May 2015)

#9  There's no outlet for the waste products after the hunny is sucked from the flowers  ?


And a fun fact from wiki



> To produce about 500 g of honey, foraging honey bees have to travel the equivalent of three times around the world


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## bellenuit (20 May 2015)

galumay said:


> The implication is that in opening the remaining box that he knows for sure doesnt have the pardon, the other remaining box *does* contain the pardon, so switching boxes would appear to increase the chance of freedom!




That wasn't implied. The two boxes that you don't pick will either have one containing the pardon or none containing the pardon. So there will always be at least one of those two that doesn't contain the pardon. The judge opens one of the two that *HE* knows doesn't contain the pardon and shows you it. The other may or may not contain the pardon as  is also the case for the one you pointed to.

If what you said was the implication was correct, then it would mean that you would never point initially to the right box, which is not the case.


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## galumay (20 May 2015)

bellenuit said:


> *
> 
> Assuming you are told the number of blue and the number of red balls in the box at the outset, can you determine the colour of the last remaining ball. You can assume there are at least two balls in the box to begin with.*




Yes, both balls are even in number, ie 4 red 2 blue, then you will end up with a red ball. both uneven you will end up with a blue ball, if one colour is uneven and one even, you will end up with the ball that is the same colour as the balls of the uneven number. ( so if 3 red, 2 blue you will end up with 1 red ball)


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## galumay (20 May 2015)

bellenuit said:


> That wasn't implied. The two boxes that you don't pick will either have one containing the pardon or none containing the pardon. So there will always be at least one of those two that doesn't contain the pardon. The judge opens one of the two that *HE* knows doesn't contain the pardon and shows you it. The other may or may not contain the pardon as  is also the case for the one you pointed to.
> 
> If what you said was the implication was correct, then it would mean that you would never point initially to the right box, which is not the case.




I think you misunderstand what i said - and I cant understand what you are saying now!

The judge knows the contents of all the boxes.

He knows the box he opens doesnt contain the pardon. Given that we already knew that he knows the contents of all the boxes that implies that the other unopened box contains the pardon in this case.

It doesnt mean i cant pick the box with the pardon, just that i didnt this time.

If you think the answer is that it doenst improve my chance of freedom by choosing the remaining box, then i struggle to see the point of the puzzle. Obviously if the judges actions provide no new information then its 50/50 which box the pardon is in.

(maybe this one is just a really basic and simple puzzle and i am overthinking it??!!)


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## cynic (20 May 2015)

Initially there is a 2/3 chance of being wrong.

After the judge reveals one of the chosen boxes to be empty one is faced with a better chance of being right by changing over to the remaining box.


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## keithj (20 May 2015)

galumay said:


> If you think the answer is that it doenst improve my chance of freedom by choosing the remaining box, then i struggle to see the point of the puzzle.



Could you explain it with 10 boxes ?  You pick one, and then the judge reveals 8 of the others are empty.  Will the odds change ?


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## galumay (20 May 2015)

cynic said:


> Initially there is a 2/3 chance of being wrong.
> 
> After the judge reveals one of the chosen boxes to be empty one is faced with a better chance of being right by changing over to the remaining box.




I dont get your logic, i agree initially there is a 33% chance of your pick being right, but once the judge reveals the empty box there is now a 50% chance you have the right box, changing to the remaining one will mean you still have a 50% chance of being correct.


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## galumay (20 May 2015)

keithj said:


> Could you explain it with 10 boxes ?  You pick one, and then the judge reveals 8 of the others are empty.  Will the odds change ?




The odds change of either of the remaining boxes being the path to freedom, but the odds dont change by swapping your choice from one to the other remaining box.


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## cynic (20 May 2015)

galumay said:


> I dont get your logic, i agree initially there is a 33% chance of your pick being right, but once the judge reveals the empty box there is now a 50% chance you have the right box, changing to the remaining one will mean you still have a 50% chance of being correct.




No. If your first choice is wrong (2/3 likely) it will dictate the choice by the judge having to reveal the only remaining empty box indicating that there is a 2/3 chance that the unchosen box contains the pardon.


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## cynic (20 May 2015)

When the last two marbles are selected a coloured marble will be returned according to the rules. It will be the last remaining marble and it's colour will be known.

Edit: If all marbles are red at outset then last marble will be red, otherwise last marble will be blue.


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## galumay (20 May 2015)

cynic said:


> No. If your first choice is wrong (2/3 likely) it will dictate the choice by the judge having to reveal the only remaining empty box indicating that there is a 2/3 chance that the unchosen box contains the pardon.




Huh?  As i said, my first choice has a 33% chance of being correct, once the judge reveals one of the remaining two boxes to be empty, there is a 50% chance of either of the two remaining boxes containing the pardon. 

Changing the box I have selected at this point will not change the odds from 50%.


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## galumay (20 May 2015)

cynic said:


> When the last two marbles are selected a coloured marble will be returned according to the rules. It will be the last remaining marble and it's colour will be known.
> 
> Edit: If all marbles are red at outset then last marble will be red, otherwise last marble will be blue.




Its more complex than that, see my answer on the previous page. Reread the question too, he is asking if you can tell upon being told what number of balls are present - eg if their are 87 red balls and 46 blue, can you tell what the remaining ball will be? (it will be red)


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## skyQuake (20 May 2015)

galumay said:


> I think you misunderstand what i said - and I cant understand what you are saying now!
> 
> The judge knows the contents of all the boxes.
> 
> ...





Wiki monty hall. They even made a US game show out of it!

Every time I have to work hard to understand it, then I get it. 6 months later I forget the reasoning again...


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## skyQuake (20 May 2015)

keithj said:


> #7 make a snowball containing the $$ & throw it back over ?




Correct



keithj said:


> #10.  He's not wearing bowling shoes.... everyone else is.




Correct



galumay said:


> #8 the alarm clock in the bookshelf




Correct



> #11 elephant has a bandage on its trunk!




Thats no elephant! But close




keithj said:


> #9  There's no outlet for the waste products after the hunny is sucked from the flowers  ?
> 
> 
> And a fun fact from wiki




Not quite but the wiki article is a huge clue



> Answer to # 12 : Use the wrench to remove car tyre tubes and tie them to the bag of coins with the rope.




Correct!


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## cynic (20 May 2015)

galumay said:


> Huh?  As i said, my first choice has a 33% chance of being correct, once the judge reveals one of the remaining two boxes to be empty, there is a 50% chance of either of the two remaining boxes containing the pardon.
> 
> Changing the box I have selected at this point will not change the odds from 50%.




There is only a 1/3 chance of you being right by remaining with your first choice which means you have a 2/3 chance of being right by changing it to the remaining box!!!


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## cynic (20 May 2015)

galumay said:


> Its more complex than that, see my answer on the previous page. Reread the question too, he is asking if you can tell upon being told what number of balls are present - eg if their are 87 red balls and 46 blue, can you tell what the remaining ball will be? (it will be red)




Okay I see your point.

If there is an uneven number of blue balls present then the result will end in a blue ball, otherwise it will be red.


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## skc (20 May 2015)

#7 - The river is frozen. Just walk across.

#8 - Cassandra Cat is hot! The libralian is clearly a geek. How hard can it be?

#9 - The company is listed on ASIC scam watch

#10 - No one wears a double breasted suit to bowl.

#11 - What animal is that in the middle? Ant eater?

#12 - Inflatable bags (inflated with the scuba regulator) are often carried by divers. 

#13 - There's a pair of eyes hiding under the machine.

#14 - Are these cartoons from a left- or right-hand drive country?

#15 - The make/model/colour of the vehicle as supplied by one of the 6 eyewitnesses.


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## keithj (20 May 2015)

cynic said:


> If there is an uneven number of blue balls present then the result will end in a blue ball, otherwise it will be red.




And the reasoning is...
There are 3 possible outcomes of a 'turn'
red,red    -2R +1R    nett result is one less red, but same number of blues
blue,blue  -2B +1R    nett result is one more red, but 2 fewer blues
red,blue   -1R -1B +1B   nett result is one less red, but same number of blues

So count of blues can only ever go down by 2 or 0, but reds can go up or down by exactly 1
Therefore blues that start with an odd number must end with an odd number (ie 1)


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## bellenuit (20 May 2015)

cynic said:


> Okay I see your point.
> 
> If there is an uneven number of blue balls present then the result will end in a blue ball, otherwise it will be red.




I can't remember all answers given, but this is the correct one and some others may have got it too.

*An even number of blues at the start means you end with just one red and an odd number of blues means you end with just one blue. You actually don't need to know the number of reds at the start.
*
The logic is as follows:

a. If 2 blue balls are picked, that will result in the box changing by -2 blue, + 1 red (2 blues are taken out, 1 red put back in)
b. If 2 red balls are picked, that will result in the box changing by -1 red (2 reds are taken out and 1 red put back in)
c. If 1 red and 1 blue are picked, that will result in the box changing by -1 red (1 red and 1 blue taken out and 1 blue put back in)

So no matter what is picked, the number of blue balls either doesn't change at all or changes by -2. 

So if you start with an even number of blue balls, they will eventually be removed going 6, 4, 2, 0. When down to 2 blues, they will be either picked together and replaced by a red (a. above) so that all remaining balls are red and thus will result in -1 red each subsequent draw (b. above) until there is no blue and just 1 red or they will be picked individually with a red (c. above) but not depleted. If this keeps happening you will eventually have just the 2 blues, which will be removed and replaced by a red. So an even number of blues result in a red.

An odd number of blues will eventually go 7, 5, 3, 1. If that remaining 1 blue is on its own, that is the game end. If there are still reds and that 1 blue is picked with a red (c. above), then it keeps getting replaced so that there is still one blue, but reds are depleting by 1 each draw. Eventually you will just have 1 blue and 1 red, which when drawn results in a blue.

So an even number of blue, means you end in a red and an odd number means you end in a blue. You actually don't need to know the number of reds at the start.


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## bellenuit (20 May 2015)

keithj said:


> And the reasoning is...
> There are 3 possible outcomes of a 'turn'
> red,red    -2R +1R    nett result is one less red, but same number of blues
> blue,blue  -2B +1R    nett result is one more red, but 2 fewer blues
> ...




Sorry, keithj, I didn't see your explanation until I had made mine.


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## bellenuit (20 May 2015)

*You are a criminal found guilty of a serious crime and have been sentenced to death. The judge, being a puzzle freak, decides to give you a chance of avoiding death.

He shows you three closed boxes on a table. You are told two are empty, but the third contains a royal pardon. You are to point to one of the boxes and if it is the one with the pardon inside you will be freed, otherwise you will be put to death.

You point to one of the boxes. The judge, knowing which box contains the pardon, opens the lid of one of the remaining boxes, one which he knows for sure doesn't have the pardon and shows you that it is empty.

He then tells you that he will allow you to change your mind if you want. You can either stick with the box you originally pointed to or you can pick the other unopened one instead. Assuming you do not want to die, does changing you choice increase your chance of freedom or make no difference. Explain why.*

Yes as someone pointed out this is the famous Monty Hall problem that caused a lot of disputes between mathematicians since it was first postulated. Only the diehards do not consent that you increase your odds by switching to the other box.

The logic is quite simple and I can't really see why it would be disputable. Your chances of being right in the first pick is 1/3 and your chances of being wrong 2/3. Because the judge pointed out which of the other two didn't have a pardon, your original choice still remains at 1/3, but the unopened box now gets the combined probability that was previously attached to the other 2, namely 2/3. So your chances of being pardoned increases if you switch.

As someone pointed out, think if there were 100 boxes with only 1 pardon. You pick one and the judge opens 98 of the remaining 99, all of which he knows are empty. To assume your probability of *not switching* jumps from 1/100 to 1/2 is absurd. The other unopened box now carries the probability that the combined 99 previously carried, namely 99/100, so switching is the thing to do.


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## galumay (20 May 2015)

bellenuit said:


> *Y
> As someone pointed out, think if there were 100 boxes with only 1 pardon. You pick one and the judge opens 98 of the remaining 99, all of which he knows are empty. To assume your probability of not switching jumps from 1/100 to 1/2 is absurd. The other unopened box now carries the probability that the combined 99 previously carried, namely 99/100, so switching is the thing to do.*



*

Your whole explanation makes no sense to me! Not saying its wrong, just that I still dont get it!!

If i have picked 1 box out of 100 i have a 1% chance of being correct, if the judge opens 98 of the 99 left and proves they all have nothing in them, we now have 2 boxes, 1 with the pardon, 1 without. Its patently obvious that which ever box you choose has a 50% chance of containing the pardon, so switching cant change your odds.

Your reasoning makes not the slightest sense to my dumb head!!

OK, read the wiki and all i can say is i am happy to be wrong along with so many geniuses and mathematicians! Still dont believe it though.*


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## skyQuake (20 May 2015)

galumay said:


> Your whole explanation makes no sense to me! Not saying its wrong, just that I still dont get it!!
> 
> If i have picked 1 box out of 100 i have a 1% chance of being correct, if the judge opens 98 of the 99 left and proves they all have nothing in them, we now have 2 boxes, 1 with the pardon, 1 without. Its patently obvious that which ever box you choose has a 50% chance of containing the pardon, so switching cant change your odds.
> 
> ...




Here's a simple explanation:
Imagine now instead that instead of revealing a box,
The judge lets you 
a) keep your initial choice
b) open BOTH remaining boxes.

b) is now a no brainer (switching) 66% vs 33%

b) is also the same as giving you both boxes (with one already open)


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## skyQuake (20 May 2015)

Correct:
7 - Throw money across in a snowball, walking on thin ice is too dangerous
8 - Alarm clock on shelf
10 - He's not wearing bowling shoes.... everyone else is.
11 - Anteater has no teeth!
12 - Use the wrench to remove car tyre tubes and tie them to the bag of coins with the rope. 

Unanswered:
9. Wiki gives clue
13. Again a general (actually fairly specific) knowledge q
14. Left/Right hand drive not actually relevant
15. Look for visual clues


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## galumay (20 May 2015)

9.) the official answer is very poor! "Honey doesnt come from flowers". I prefer the fact that there is no waste stream, once the pollen has been converted to honey the rest of the flower should come out as waste. Not one of his best!

13.) rubber shrinks when you heat it.

14.) it happened "here" but no glass on road?

15.) brown car with concrete splashed on it, blue paint from car & letter box on it and panel damage!


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## bellenuit (21 May 2015)

galumay said:


> Your whole explanation makes no sense to me! Not saying its wrong, just that I still dont get it!!
> 
> If i have picked 1 box out of 100 i have a 1% chance of being correct, if the judge opens 98 of the 99 left and proves they all have nothing in them, we now have 2 boxes, 1 with the pardon, 1 without. Its patently obvious that which ever box you choose has a 50% chance of containing the pardon, so switching cant change your odds.
> 
> ...




Just because there are two choices doesn't mean each choice is of equal probability.

Think of it this way. If you believe that staying with your first choice gives you a 50/50 chance of picking the right box, then if you were to play that game 1000 times, you are saying that you would expect to pick the correct box out of the 100 five hundred times on average. You know each time you play that at least 98 of the other 99 boxes are empty. Whether the judge opens 98 of those boxes which HE knows before hand are empty and shows them to you or he just leaves them all closed, why would your chances of winning be different? 

I think you are confusing this problem with randomly opening the other boxes and each time one is found without a pardon then the probability of all the rest (including yours) having the pardon increases. By the time you have opened all but one of the remaining boxes and still haven't found one with a pardon, then obviously yours and that box is down to a 50/50 chance. Every time a box was opened at random, it had a 1/100 chance of having the pardon, and if it had, then that was it. What was left would then all have had zero chance. If it didn't have the pardon, all the remaining boxes had an increased chance. 

But this problem is different. The judge knows which boxes do not have a pardon and opens only those. Unlike the other scenario there was never a possibility that opening one of those boxes could have revealed the pardon, so there was never a possibility of the remaining boxes suddenly having a zero chance.




Each of those boxes have a 99 out of 100 chance of


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## keithj (21 May 2015)

skyQuake said:


> Here's a simple explanation:
> Imagine now instead that instead of revealing a box,
> The judge lets you
> a) keep your initial choice
> ...




Here's 3 other ways to think about it....

1) You could try the experiment 100 times & see how many times you're successful by switching.

2) What if you used a pack of cards - you have a 1/52 chance of picking Ace of Spades (& a 51/52 chance of it being elsewhere) .  Get your butler to sneak a look at your card - if it's the Ace then he should remove 50 of the remaining cards, otherwise just leave the Ace there by itself for you to optionally choose.

It should be clear that switching will have a 51/52 chance of being correct.

Try that 100 times too!

3) Think of it as 2 separate events. The 1st event has a 1/52 chance of success. The 2nd event (ie the optional switch) has a 100% chance of success (but only if the 1st event proved unsuccessful). So the odds of success by switching are 51/52 x 100%.  While the odds of success by NOT switching will remain at 1/52.


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## galumay (21 May 2015)

bellenuit said:


> How many have guessed 3 boxes and how many have guessed all 4 boxes correctly?




If they guessed 3 correctly, then they know the fourth as well, so 10 picked all 4 boxes. (and all 3 boxes by definition.)


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## bellenuit (21 May 2015)

galumay said:


> If they guessed 3 correctly, then they know the fourth as well, so 10 picked all 4 boxes. (and all 3 boxes by definition.)




Yes, or to put it another way, it is impossible to just guess exactly 3 correctly, so 0 guessed and the remaining 10 guessed all 4. Too easy.


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## pixel (21 May 2015)

bellenuit said:


> Yes, or to put it another way, it is impossible to just guess exactly 3 correctly, so 0 guessed and the remaining 10 guessed all 4. Too easy.




sorry Belle,
I hadn't seen this as the solution thread. Shall put my guesses here in future.


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## pixel (21 May 2015)

> You, a Greek mathematician, are swimming in a perfectly round lake when you see a savage cyclops eying you from the shore. He has manoeuvred to the closest point on the shore to where you currently are and there is nothing he would like more than to eat you. You know the cyclops can't swim, but he can run 4 times faster than you can swim. However, you can run much faster than the cyclops, so you know if you can get to the shore before he gets to you, you can outrun him and escape.
> 
> You are a good swimmer and can maintain your maximum speed indefinitely even if it involves sharp changes in direction (e.g. no loss of speed if you switch direction). The same applies to the cyclops. He can run 4 times faster than you can swim and switch direction without loss of pace.
> 
> Is it possible for you to get to the shore and escape assuming he always tries to get to the point on the shore where you are trying to swim to? If so, explain how.



Answer: I don't think it's possible.
Start in the middle of the lake and swim away from where the Cyclops stands. That will give you the shortest distance to shore, which is the farthest away from Cyclops. He will have to run a half circle, the length of which is 3.1415... times the radius that you need to swim. But he runs 4-times as fast as you swim.


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## skc (21 May 2015)

bellenuit said:


> Is it possible for you to get to the shore and escape assuming he always tries to get to the point on the shore where you are trying to swim to? If so, explain how.[/B]




So say I am at near 12 o'clock position of the round lake, but quite close to the shore... if I turn around and head towards the 6 o'clock shore, does the cyclop run towards that point?

If so, can I not simply swim slower? Say I swim at a speed such that I am only 1/4 of the way across the lake when the cyclop has already made it to the opposite shore. At which time I simply swim at maximum speed back to the 12 o'clock edge. I only need to swim a distance of 0.5R when the cyclop needs to cover 3.14159R, or >6x as far.


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## pixel (22 May 2015)

skc said:


> So say I am at near 12 o'clock position of the round lake, but quite close to the shore... if I turn around and head towards the 6 o'clock shore, does the cyclop run towards that point?
> 
> If so, can I not simply swim slower? Say I swim at a speed such that I am only 1/4 of the way across the lake when the cyclop has already made it to the opposite shore. At which time I simply swim at maximum speed back to the 12 o'clock edge. I only need to swim a distance of 0.5R when the cyclop needs to cover 3.14159R, or >6x as far.




yes, that should work. Provided the cyclops is stupid enough to run at maximum speed to that far 6 o'clock point before I even made it half-way to the middle of the lake.

A smart cyclops would only need to stay on the side of the lake and move to the point that you're closest to, never overtake you.


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## galumay (22 May 2015)

pixel said:


> Answer: I don't think it's possible.




Agreed, as long as he is a smart cyclops!


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## bellenuit (22 May 2015)

skc said:


> So say I am at near 12 o'clock position of the round lake, but quite close to the shore... if I turn around and head towards the 6 o'clock shore, does the cyclop run towards that point?
> 
> If so, can I not simply swim slower? Say I swim at a speed such that I am only 1/4 of the way across the lake when the cyclop has already made it to the opposite shore. At which time I simply swim at maximum speed back to the 12 o'clock edge. I only need to swim a distance of 0.5R when the cyclop needs to cover 3.14159R, or >6x as far.




Sorry, that was bad phrasing on my behalf. He will only try to get to the point you are heading to if you are swimming away from the centre. It's hard to think of all contingencies. Just assume he is smart and will always stay as close or try to get as close on the shoreline as he possibly can to where you currently are. So if you are heading towards the centre, he will stay or head to the point if not already there that is closest to where you currently are in case you backtrack. If heading away from the centre he will try and get to the point that you seem to be heading to. I'll add something to the original to make this clearer.


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## bellenuit (22 May 2015)

I'll post the answer about midday EST Friday.


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## galumay (22 May 2015)

mmm...i have been doing some fiddling with my basic maths and a ruler, compass and pencil!

Seems to me there is a circle that I can swim in where I will be able to move faster than the cyclops relative to the circle he has to run around. That means i can get opposite him, while closer to the bank than the center.

The bit of maths i cant quite get my head around is whether that circle will mean i am close enough to the bank to make it before the cyclops - i suspect it is, but i cant be certain.

I need to make a sprint to the edge of less than 3.89m to beat him to the bank if he is directly opposite. (based on a lake of 10m diameter).


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## cynic (22 May 2015)

Presuming that I start in the centre of the lake. I'd first swim PI/8 lake radius in the opposite direction to the Cyclops. In that time the Cyclops will have covered 1/4 lake circumference in his efforts to draw closer. At that point I'd change direction and swim PI/8 radius towards the shore opposite Cyclops new position. Again he'd cover 1/4 circumference in that time. I'd continue this exercise until such time as the distance between myself and the shore opposite the Cyclops position was slightly less than PI/8 lake radius, at which point I'd safely swim to shore knowing that the Cyclops won't be able to reach me in time to catch me.


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## Tisme (22 May 2015)

Problem #168/#169

Cyclops has the specific advantage of 4/ ¶. So the swimmer has to swim >4/ ¶ away from the Cyclops. I'm guessing that because the swimmer doesn't have his trig tables nor a handy calculus ready reckoner he's going to go three dimensional and duckdive into the water and swim far enough (away from Cyclops) concealed to reduce the distance to the shoreline to < ¶r/4....?


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## keithj (22 May 2015)

1. Swim to centre of lake
2. swim in a straight line directly away from cyclops
3. when cyclops moves towards closest point on shore swim slightly more than 180 degrees away from him. This change in direction should be sufficient to force cyclops to change direction.
4. Repeat step 3 until you have swum 1/4 distance to shore.  This zigzagging forces cyclops to oscillate about the point furthest away from you. At the 1/4 radius distance the angular speed you can swim is the same as the speed cyclops can run, so any more zigzagging will prove pointless
5. Swim directly to shore safe in the knowledge that you can swim 3/4 of the radius faster than he can run 1/2 the circumference.

As an alternative to steps 3 & 4. Simply swim out to a distance slightly less than 1/4 of the radius and then swim in a circle until you are directly opposite cyclops, and then do step 5.  

These methods work because your angular speed is greater than cyclops angular speed if you remain inside the 1/4 radius and consequently you can get opposite him.


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## bellenuit (22 May 2015)

Both Cynic and Keith may be right, but I would like to see the math fully worked out so that we can be 100% certain. I think in both cases they will have swum into the area between the two inner circles in the answer below, which then allows them to get out safely.

The answer is he can get out safely. This is one proof, but as I said, zig zag methods may also prove successful.





Looking at the three circles in the diagram, the outer circle represents the lake. It has radius R metres. Lets assume you can swim at speed V metres per second.

We know that if you were at the centre of the lake and swam directly to the side, the cyclops could always get there before you, because it would take you R/V seconds to get to the shore and the maximum time he needs to run around half the circumference of the lake is πR/4V seconds. Since π is 3.14 approx, then π/4 is less than 1, hence the cyclops will always get there first.

In the time the cyclops takes to run a half circumference, πR/4V seconds, you can swim πR/4 metres. This is 0.79R and is represented by the distance D in the diagram (not to scale). So it is clear if you can manage to to be anywhere less than D metres from the shore when the cyclops is at the exact opposite side of the lake (180 degrees removed), you can get to the shore before he can get to your point of exit. The innermost circle of radius A represents all points where *you would NOT make it* to the shore  before the cyclops if he were to start 180 degrees removed from you. A is obviously R-D, or 0.21R. So if you are anywhere outside that innermost circle and the cyclops is 180 degrees removed from you, you can get to the shore before he gets to you. So how do you manage to do this?

The time taken for the cyclops to fully complete a circle is 2πR/4V or πR/2V seconds. The circle with radius B represent the circle that you can fully swim around in the same time it takes the cyclops to run around the lake. The time you take to swim around that circle is 2πB/V seconds and if this is the same time as it takes the cyclops to run around the lake, we get:

2πB/V = πR/2V or B = R/4 or 0.25R

So if you are on that circle with radius B, you can swim around it in the same time it takes the cyclops to run around the lake. Obviously if you are on a smaller circle (one with radius less than B) you can swim around that smaller circle faster than the cyclops can run around the lake. 

Although I have drawn it that way, it wasn’t known until we worked the above out that the circle with radius A (where A = 0.21R) is inside the circle with radius B (where B is 0.25R). This is fortunate as it means that if we swim to anywhere in the area between the two inner circles, because we are within the circle with radius B we can swim in a circular fashion around the lake faster than the cyclops can run around it, so we will be able to get to a point where he is 180 degrees removed from where we are, but because we are outside the circle with radius A, we can then get to the shore before he can get around to the point we exit.

So to escape, you just need swim to just outside the A circle and then swim on an imaginary circle around the centre until we are 180 degrees removed from where the cyclops is. At that point we then swim directly to the shore and will be able to get out before he makes back around.


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## skyQuake (22 May 2015)

Just swim away from him in the opposite direction he is facing.

Being a cyclops he lacks depth perception and thus you'll get to at least 0.5 radius before he notices. By then its too late


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## skc (22 May 2015)

skyQuake said:


> Last Slylock I got.




a). The kids knew the cookies were Oatmeal?

b). Tilt the drum and pour it until the fluid surface makes a perfect diagonal across the height of the drum

c). The message would take 5 million years to reach the aliens. Although, it's questionable what speed "intergalactic radio message" travels at.

d). A 3 prong plug on the computer vs 2-hole power point from the wall. 



skyQuake said:


> If you enjoyed, I highly recommend buying the books. For kids/partner/yourself(?)




I have to say that the questions are not logically watertight enough for an adult... but I'd love it as a kid. I have a friend's 8yr old boy birthday coming up...


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## keithj (22 May 2015)

bellenuit said:


> This is one proof, but as I said, zig zag methods may also prove successful.



I think the proof is simply that the angular velocity of the swimmer is always greater than the cyclops provided he stays within 1/4 of the radius, and he can afford to swim slightly further out with each zig. The swimmer would have to consult trig tables to calculate exactly what angle is optimum at each zig & zag, but in practice the swimmer would just swim more parallel with the shore until the cyclops reversed.

The zigzags would get shallower as the distance from the centre increased. On the whole I think the zigzag method would be more efficient than the swimming in a circle method (and it would have the benefit of p***ing the cyclops off ).

And many thanks to you & others for providing these puzzles .


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## skyQuake (22 May 2015)

skc said:


> a). The kids knew the cookies were Oatmeal?
> 
> b). Tilt the drum and pour it until the fluid surface makes a perfect diagonal across the height of the drum
> 
> ...




Well done. All solved.

Let me know if the kid can answer the questions about displacement (finger in glass of water q), rubber shrinking when heated etc!


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## bellenuit (22 May 2015)

keithj said:


> I think the proof is simply that the angular velocity of the swimmer is always greater than the cyclops provided he stays within 1/4 of the radius, and he can afford to swim slightly further out with each zig. The swimmer would have to consult trig tables to calculate exactly what angle is optimum at each zig & zag, but in practice the swimmer would just swim more parallel with the shore until the cyclops reversed.
> 
> The zigzags would get shallower as the distance from the centre increased. On the whole I think the zigzag method would be more efficient than the swimming in a circle method (and it would have the benefit of p***ing the cyclops off ).
> 
> And many thanks to you & others for providing these puzzles .




Actually the quickest way to get out of the lake safely is to swim *in a spiral* to a point on the safety circle (the innermost circle with radius A) according to the source article, but the mathematics to prove that is beyond me. Once at the safety zone, one obviously then swims directly to shore.

The spiral makes sense. If you swim directly to the safety circle, the cyclops will have moved to the point directly in line to where you are heading and you then have to swim in a circle until you get 180 degrees ahead of him. Swimming in a spiral means the cyclops will not be at the point directly nearest the point you touch the safety circle, so you will not have to swim as much in a circular direction.

The spiral is like a continuous changing zig, without the zag. Zagging doesn't make sense IMO. Once the cyclops starts running round the lake in anticipation of where you are heading, it never then makes sense for him to change direction, hence causing you to change direction and zag. Once the cyclops has decreased the angular distance between you and him to under 180 degrees (say to 160 degrees), for him to switch direction would suddenly mean the angular separation is now 200 degrees based on the direction he is now heading if you were then to zag, leaving him worse off than if he kept going in the same direction.


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## bellenuit (22 May 2015)

_Actually the quickest way to get out of the lake safely is to swim in a spiral to a point on the safety circle (the innermost circle with radius A) according to the source article, but the mathematics to prove that is beyond me. Once at the safety zone, one obviously then swims directly to shore._

I need to change that on re-reading the article. Spiralling is the fastest way to get out, but it doesn't say explicitly that you spiral just to the innermost circle. It just says you spiral to somewhere between the two inner circles and then you still must circle until you are 180 degrees ahead of him before making a direct exit.


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## cynic (25 May 2015)

According to my calculations Denise was born 14th May 2002.


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## bellenuit (25 May 2015)

cynic said:


> According to my calculations Denise was born 14th May 2002.




This is my answer too. I reckon it is easier than the original Cheryl's Birthday.

My workings for those who are having difficulty (the number in brackets beside the dates indicate the statement below it that eliminated that particular date)

17 Feb 2001 (3), 16 Mar 2002 (6), 13 Jan 2003 (4), 19 Jan 2004 (5)

13 Mar 2001 (4), 15 Apr 2002 (2), 16 Feb 2003 (7), 18 Feb 2004 (5)

13 Apr 2001 (2), 14 May 2002, 14 Mar 2003 (6), 19 May 2004 (5)

15 May 2001 (3), 12 Jun 2002 (1), 11 Apr 2003 (1), 14 Jul 2004 (5)

17 Jun 2001 (2), 16 Aug 2002 (7), 16 Jul 2003 (7), 18 Aug 2004 (5)

Denise then told Albert, Bernard and Cheryl separately the month, the day and the year of her birthday respectively.

The following conversation ensues:

*Albert: I don’t know when Denise’s birthday is, but I know that 
Bernard does not know.*
_1. This rules out 12 Jun 2002 and 11 Apr 2003 because these two DAYs are unique. 
2. But the only way Albert can be sure that it is not these two days is because he knows the MONTH is not Apr or Jun. So that also rules out the remaining dates with those months: 15 Apr 2002, 13 Apr 2001 and 17 Jun 2001.  _

*Bernard: I still don’t know when Denise’s birthday is, but I know that Cheryl still does not know.*
_3. Since Bernard knows the DAY, this rules out 17 Feb 2001 and 15 May 2001, as these DAYS are unique from what is left over. 
4. Because he knows that Cheryl still doesn’t know the birthday, he must know the DAY isn’t 13, because 13 Mar 2001 is the last left over DAY in 2001 meaning Cheryl could know the answer if it were that YEAR. So this eliminates 13 Mar 2001 and also the other DAY with 13: 13 Jan 2003._

*Cheryl: I still don’t know when Denise’s birthday is, but I know that Albert still does not know.*
_5. Since Cheryl knows the YEAR, she must know that it isn’t 2004 because that has a date with a unique left over MONTH, 19 Jan 2004, so if it were 2004, Albert who knows the month could know the birthday. This eliminates all 2004 dates: 19 Jan 2004, 18 Feb 2005, 19 May 2004, 14 Jul 2004 and 18 Aug 2004._

*Albert: Now I know when Denise’s birthday is.*
_6 Since Albert knows the MONTH and knows the answer, then it must be one of those leftover dates that has a unique MONTH. This eliminates the two dates that do not have a unique MONTH: 16 Mar 2002 and 14 Mar 2003._

*Bernard: Now I know too.*
_7. Since Bernard knows the DAY and now knows the birthday, then it must be the 14 May 2002, as the remaining dates have a common DAY, 16. So 16 Feb 2003, 16 Aug 2002 and 16 Jul 2003 are eliminated. The only date left is 14 May 2002, Denise’s birthday. 
_
*Cheryl: Me too.*


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## luutzu (27 May 2015)

Saudi:

The wise man told them that the Sheik meant to race the camels once it arrives? So the slowest camel will be the one most tired?


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## luutzu (28 May 2015)

Billiard:

1. Put 3 on each side - if it balances, then go to step 2.

2. Put the 6 you just scaled aside, divide the other 6 into 2 group * of 3 [say, B1 and B2].

2.1 Take B1 and scale against A1 [any 3 balls, all 6 are identical as they balanced]

If B1 and A1 balanced, then the "defective" one must be in B2... 

brb...*


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## bellenuit (28 May 2015)

luutzu said:


> Saudi:
> 
> The wise man told them that the Sheik meant to race the camels once it arrives? So the slowest camel will be the one most tired?




This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.


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## bellenuit (28 May 2015)

luutzu said:


> Billiard:
> 
> 1. Put 3 on each side - if it balances, then go to step 2.
> 
> ...



*

You have to find the actual ball, not just a group that it is in and you have to determine whether it is heavier or lighter than the others, not just that it is different (defective).*


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## luutzu (28 May 2015)

bellenuit said:


> This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.




Yea. I googled the answer.. haha... and didn't agree with their solution 
Mine above is bad too... but swapping? That's assuming the other prince allow the swap, and also assumes they've been riding around for days on a few camels, one or two of which happen to belong to other princes.


Haven't finish the billiard yet. Havng too much fun with my new toy at the moment


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## keithj (30 May 2015)

bellenuit said:


> This is a question that could have many answers obviously, but I think what this question is trying to get at is how the two poor guys could avoid spending the rest of their lives in the desert and still have a chance of winning. Colluding so they pass the line at exactly the same time doesn't count.




Not sure what advice the old man gave, but jumping on each others camel & racing to city seems to be the answer. Effectively they are attempting to make the others camel  faster.


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## keithj (30 May 2015)

bellenuit said:


> *You have 12 billiard balls that look identical in every respect, except one is slightly different in weight (could be heavier or lighter than the other 11). You have a simple balance (one that stays level if both trays contain exactly the same weight, but will tip one way or the other if they are different).
> 
> You have to determine by no more than 3 weighings, which ball is different and whether it is heavier or lighter than the others. How can you do it?*




This seems to be a complicated method .....  

The balls can be in one of 4 states - Unknown, CorrectWeight, CouldBeHeavier, CouldBeLighter
All 12 balls start in the Unknown state.
If they are proven to be the same weight as another ball (or set of balls) they are labeled CorrectWeight & can be eliminated
If they are proven to heavier or lighter than other balls then they are labeled as such.

Empty the balance prior to any weighing.

```
Label all 12 balls with [B]Unknown[/B].

Weigh 1.
Put 4 on each side
If they balance then
    Label those eight on the balance as C (for Correct weight). We now have 4 Unknowns.
    Weigh 2.
         Put 3 (of the 4 remaining) Unknowns on one side the balance & 3 of the C on the other.
         If they balance then
                label the 3 Unknowns on the balance as C. 
                So we now have 1 Unknowns but we don;t know if it's heavier or lighter, so...
                Weight 3a.
                    Put the only remaining Unknowns of the balance opposite one C, so we can find out if it's heavier or lighter
         else if the C and Unknown is heavier then
            label the Unknown on the heavier side as [B]UBCBH [/B](Unknown but could be heavier)
            label the 2 Unknowns on the lighter side as [B]UBCBL [/B](Unknown but could be lighter)
            Weigh 3b
                    Put one UBCBH and one UBCBL on one side and 2 C on the other
                    if they balance then
                            the UBCBL that is not on the scales is the lighter than the other 11
                    if the side with one UBCBH and one UBCBL is heavier then the UBCBH is heavier than the other 11
                    else UBCBL is lighter than the other 11
         else if the 2 Unknowns are heavier then
                label them both UBCBH
                Weight 3c.
                    Label the 2 Unknowns that aren't on the balance as C
                    Put both the UBCBH on opposite sides of the balance
                    One will be heavier than the other 11


As they don't balance then
        label the 4 not on the balance as Correct
        label the heavier 4 Unknowns as UBCBH and the other 4 UBCBL
        Weigh 2a.
        Put 2 UBCBH and one UBCBL on the LHS of the balance, and 1 UBCBH and 2 UBCBL on the RHS
        and leave one of each aside.
        if they balance then
            Label all 6 on the balance as C
            Weight 3d.
            Put the remaining UBCBL on the balance opposite a C
            if they balance then the remaining one (ie not on the balance)  is heavier than the other 11
            else the one on the balance (labeled UBCBL) is lighter than the other 11
        else they don't balance
            Label the 2 that aren't on the balance as C
            if the LHS is heavier (it contains 2 UBCBH and only one UBCBL) then
                there are 2 possibilities (either the lighter one by itself is lighter than all others, or one of the 2 UBCBH is heavier than all others)
                Label the UBCBL on the LHS as C
                Label the 2 UBCBH on the RHS as C
                We are left with 3 balls - 1 UBCBL and 2 UBCBH
                Weight 3e.
                    Put one of the UBCBH and the only remaining UBCBL opposite 2 C
                    if they balance then the UBCBH is heavier than the other 1
                    else if the 2 Cs are heavier the the UBCBL on the balance is lighter than the other 11
                    else the UBCBH on the balance is heavier than the other 11
            else the RHS must be heavier then
                just do the inverse of Weight 3e.
```


Is there an easier way ?


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## bellenuit (30 May 2015)

keithj said:


> Not sure what advice the old man gave, but jumping on each others camel & racing to city seems to be the answer. Effectively they are attempting to make the others camel  faster.




Yes, that was the advice. Swap camels.


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## cynic (30 May 2015)

*Scenario 1:*

Weigh balls 1-4 against 5-8

If even weigh balls 1-3 against 9-11 

If even weigh ball 12 (which is oddball) against 1 to determine relative weight by direction of tilt.

*Scenario 2:*

Weigh balls 1-4 against 5-8

If even weigh balls 1-3 against 9-11 

If uneven 9-11 contains oddball and relative weight is now known from direction of tilt. 

Weigh ball 9 against  10 

If even 11 is the oddball otherwise it can be identified as either 9 or 10 by the direction  of the tilt.

*Scenario 3:*

Weigh balls 1-4 against 5-8  

If uneven weigh balls 4-7 against 8-11 

If even 1-3 contains oddball and relative weight is now known from direction of tilt in first weighing. 

Weigh ball 1 against 2 

If even ball 3 is the oddball otherwise it can be identified as either 1 or 2 from the direction  of the tilt.

*Scenario 4:*

Weigh balls 1-4 against 5-8  

If uneven weigh balls 4-7 against 8 - 11 

if uneven and tilt unchanged then oddball is either 4 or 8. 

Weigh ball 4 against 1. 

If uneven then oddball is 4 otherwise it is 8. 

Relative weight can now be deduced from the direction of the tilt in the second weighing.

*Scenario 5:*

Weigh balls 1-4 against 5-8  

If uneven weigh balls 4-7 against 8 - 11 

If uneven and tilt changed then oddball is in 5-7 and relative weight is now known by direction of tilt.

Weigh ball 5 against 6.

If even ball 7 is the oddball otherwise it can be identified as either 5 or 6 from the direction  of the tilt.


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## bellenuit (31 May 2015)

Both cynic and keithj seem correct. I can't see any flaw in their answers.


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## cynic (31 May 2015)

3211000


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## cynic (31 May 2015)

cynic said:


> 3211000




42101000


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## cynic (31 May 2015)

cynic said:


> 42101000




521001000


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## cynic (31 May 2015)

cynic said:


> 521001000




6210001000


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## cynic (31 May 2015)

cynic said:


> 6210001000




C210 0000 0000 1000


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## bellenuit (31 May 2015)

Cynic, correct on all which is self evident. 

They are referred to as self descriptive numbers (I didn't want to give them that name in the problem as it would be too easy to search the answers on Google).

More on them here. 

http://en.wikipedia.org/wiki/Self-descriptive_number


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## cynic (31 May 2015)

bellenuit said:


> Cynic, correct on all which is self evident.
> 
> They are referred to as self descriptive numbers (I didn't want to give them that name in the problem as it would be too easy to search the answers on Google).
> 
> ...




Thanks for that bellenuit.

Perhaps it's just that I'm particularly fond of numbers, but I found that problem to be by far the easiest you've presented thus far.


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## galumay (12 June 2015)

bellenuit said:


> *A bag contains one marble and there is a 50% chance that it is either Red or Blue.
> 
> A Red marble is added to the bag, the bag is shaken and a marble is removed. This removed marble is Red.
> 
> What is the probability that the remaining marble is Red?*




My initial logic is that its the same as if you had never added a marble, ie you put in a red marble and took out a red marble, therefore the remaining marble could either be red or blue - 50% chance.


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## skc (12 June 2015)

bellenuit said:


> *A bag contains one marble and there is a 50% chance that it is either Red or Blue.
> 
> A Red marble is added to the bag, the bag is shaken and a marble is removed. This removed marble is Red.
> 
> What is the probability that the remaining marble is Red?*




Nice question. The answer is 1/3. As there are only 2 marbles, it's easy to work out by just drawing a "possibility tree".

Gets a bit more complicated if there were a few more in the bag.


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## cynic (12 June 2015)

skc said:


> Nice question. The answer is 1/3. As there are only 2 marbles, it's easy to work out by just drawing a "possibility tree".
> 
> Gets a bit more complicated if there were a few more in the bag.




Funnily enough, I'm arriving at 25% and I thought I was applying the same methodology.

Edit: I think I've spotted my mistake and now agree with your answer.


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## skc (12 June 2015)

cynic said:


> Funnily enough, I'm arriving at 25% and I thought I was applying the same methodology.




You need to remove the Blue/Blue scenario... because it is not a valid one as it was given that we removed a red ball.

So there are only 3 possible "branches", of which only one ends up having a red in the bag.


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## galumay (12 June 2015)

skc said:


> Nice question. The answer is 1/3. As there are only 2 marbles, it's easy to work out by just drawing a "possibility tree".
> 
> Gets a bit more complicated if there were a few more in the bag.




Once again it seems the answer is totally counterintuitive! I have no doubt you are right - but for the life of me cant see how!

As I said, it seems to me its as if nothing has changed, there was a 50% chance of the ball being red or blue, you put one in and take out one the same colour, the remaining ball still has a 50% chance of being either colour.

I think this is the same as the similar problem that I couldnt get my head around - the one that many mathematicians and scholars agreed with my incorrect answer!


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## cynic (12 June 2015)

galumay said:


> ...
> As I said, it seems to me its as if nothing has changed, there was a 50% chance of the ball being red or blue, you put one in and take out one the same colour, the remaining ball still has a 50% chance of being either colour.
> ...




Your thinking would only be correct if you could be certain that you took out the same red that you put in.


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## cynic (12 June 2015)

skc said:


> You need to remove the Blue/Blue scenario... because it is not a valid one as it was given that we removed a red ball.
> ...



Yes! That was where I slipped up!


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## skc (12 June 2015)

galumay said:


> Once again it seems the answer is totally counterintuitive! I have no doubt you are right - but for the life of me cant see how!
> 
> As I said, it seems to me its as if nothing has changed, there was a 50% chance of the ball being red or blue, you put one in and take out one the same colour, the remaining ball still has a 50% chance of being either colour.
> 
> I think this is the same as the similar problem that I couldnt get my head around - the one that many mathematicians and scholars agreed with my incorrect answer!




Hang on a second... I am wrong! I forgot the bold part. 

P.S. My daughter is sick so I only got 4 hour or so sleep last night... good enough excuse?

Then it should be 50/50?!



bellenuit said:


> A bag contains one marble and there is a 50% chance that it is either Red or Blue.
> 
> *A Red marble is added to the bag, *the bag is shaken and a marble is removed. This removed marble is Red.
> 
> What is the probability that the remaining marble is Red?


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## cynic (12 June 2015)

skc said:


> Hang on a second... I am wrong! I forgot the bold part.
> 
> P.S. My daughter is sick so I only got 4 hour or so sleep last night... good enough excuse?
> 
> Then it should be 50/50?!




Now I definitely do not agree with you, unless by "Red" as opposed to "red" it means the same marble was removed.


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## bellenuit (12 June 2015)

Actually the answer is 2/3.

Let's name the initial marbles R and B and the added marble Ra

There is a 50% chance at the beginning that the bag contains R and a 50% chance that it contains B.

After Ra is added, there is now a 50% chance that the bag contains R + Ra and a 50% chance that it contains B + Ra.

A marble is drawn from the bag (at this stage assume any colour). So there are now 4 possibilities for what's inside the bag and outside the bag.

A. If the bag contained R + Ra (overall a 50% chance), then you could have:
A1. R inside and Ra outside, or
A2. Ra inside and R outside. 

There is a 50% chance for either of these outcomes (if A is true) and since there is just a 50% chance for A being true, then there is a 25% chance for A1 and a 25% chance for A2 overall.

B. If the bag contained B + Ra (overall a 50% chance), then you could have:
B1. B inside and Ra outside or
B2. Ra inside and B outside.

As before there is a 50% chance for either of these outcomes (if B is true) and since there is just a 50% chance for B being true, then there is a 25% chance for B1 and a 25% chance for B2 overall.

So there are in total four possible outcomes, each with an equal 25% probability (giving us 100%).

But since we are only considering the situation where a Red is drawn first, then we are left with just 3 possible outcomes, each with equal probability: A1, A2 and B1. Two of these have a Red left in the bag (A1 and A2) and the other a Blue left in the bag (B1), so the probability of a Red marble being left in the bag is 2/3 or 66%.


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## cynic (12 June 2015)

bellenuit said:


> Actually the answer is 2/3.




This actually matches the intended outcome of SKC's original calculation (and my revised one). It's been a pretty wild week.


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## bellenuit (12 June 2015)

cynic said:


> This actually matches the intended outcome of SKC's original calculation (and my revised one). It's been a pretty wild week.




I thought your answers might have been just an inadvertent mix up of the colours, but as there was no detailed workout given, I couldn't tell.

That puzzle is attributable to Lewis Carroll.


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## keithj (13 June 2015)

bellenuit said:


> *You have a perfect solid sphere upon which you designate two opposite points as poles (like the earth's North and South poles). You drill a perfect cylindrical hole from one pole to the other (e.g. the hole goes right through the centre). The length of the hole is 2 cm. This is the length of the edge of the hollow cylinder created by the hole, not the diameter of the sphere (or if you were to put the sphere in a vice with what were the pole ends touching the arms of the vice, this would be the separation of those arms). See image below.
> 
> What is the remaining solid volume of the sphere?
> 
> ...



The crucial hint given implies the radius of the cylinder & the radius of the sphere are not required, so ANY parameters should give the required answer. 

So by making the radius of the drilled hole infinitely small, the height of the cylinder becomes equal to the diameter of the sphere. And therefore the volume of the full sphere less the volume of an infinitely small cylinder is the answer.... (4/3 x pi x 1 x 1 x 1) - (2 x pi x 0 x 0)

However, upon checking with other examples it becomes clear that it's a bit more complex. And google quickly led to the napkin ring problem.


----------



## bellenuit (13 June 2015)

keithj said:


> The crucial hint given implies the radius of the cylinder & the radius of the sphere are not required, so ANY parameters should give the required answer.
> 
> So by making the radius of the drilled hole infinitely small, the height of the cylinder becomes equal to the diameter of the sphere. And therefore the volume of the full sphere less the volume of an infinitely small cylinder is the answer.... (4/3 x pi x 1 x 1 x 1) - (2 x pi x 0 x 0)




Correct, keithj. The remaining volume is purely dependent on the height of the cylinder, not on the radius of the sphere. As the radius of the sphere increases, the radius of the cylinder must also increase so that the height remains 2 cms. Therefore to solve the problem, you take, as you have done, the extreme case where the cylinder shrinks to zero radius, in which case the volume of the remainder is the same as the volume of the sphere itself. This sphere has a radius then of 1cm.

So the answer is: 4π/3 cubic cms (as r**3 = 1**3 = 1)

Thanks for the reference to the Wikipedia Napkin Ring problem as I would have been in strife if anyone asked for proof that the remaining volume is constant (for any given height of the cylinder) irrespective of the sphere radius.

This is the link for those who wish to pursue it and you will find a good graphic there too.

https://en.wikipedia.org/wiki/Napkin_ring_problem



keithj said:


> However, upon checking with other examples it becomes clear that it's a bit more complex.




I am confused by what you mean here. Are you saying that you have found examples where this isn't the case?


----------



## cynic (14 June 2015)

I arrived at a different answer.

Basically I calculated the result as being equal to the volume of the sphere less the volume of the cylinder and also less a volume attributable to the two removed caps at the poles.

If my trigonometry and condensation skills have served me correctly, the answer is something akin to:

pi X ((4 X R X R X R X ((1/3)-((1/90) X inverse cosine (1/R)))- (4 X ((R X R) -1)))

Edit: I'm still pondering bellenuit's explanation and am as yet undecided either way.


----------



## cynic (14 June 2015)

I calculated the radius of the cylinder (using pythagorus theorem) as being equal to sqrt[(R X R) - 1]. 

The increase in expansion of the cylinder radius (essential to maintenance of cylinder height in the event of expansion of the sphere radius) would certainly increase the volumes of the removed caps and cylinder. 

I am still undecided as to whether or not these changes exactly compensate.


----------



## cynic (14 June 2015)

Okay. 

I've looked at the Wikipedia article on the Napkin ring problem and the explanation in the proof about the volume being directly related to cylinder height regardless of sphere radius do indeed make sense. 

So I'm okay with it now, so please feel free to disregard my rather convoluted answer.


----------



## keithj (14 June 2015)

bellenuit said:


> I am confused by what you mean here. Are you saying that you have found examples where this isn't the case?



No. My initial assumption was that the formula was simply 4/3 x pi x (h/2)^3. i.e the same as a sphere for your example.  Trying an example with a non-zero hole radius soon killed that theory and hence my comment that it was a bit more complex.

The wiki page gives the correct formula of pi x h^3 / 6.

I too was heading in the same direction as  cynic (Vol of sphere - vol of cylinder - 2 x vol of end cap) when the wiki page pointed me towards napkins.


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## cynic (14 June 2015)

keithj said:


> No. My initial assumption was that the formula was simply 4/3 x pi x (h/2)^3. i.e the same as a sphere for your example.  Trying an example with a non-zero hole radius soon killed that theory and hence my comment that it was a bit more complex.
> 
> The wiki page gives the correct formula of pi x h^3 / 6.
> 
> I too was heading in the same direction as  cynic (Vol of sphere - vol of cylinder - 2 x vol of end cap) when the wiki page pointed me towards napkins.




I'm not sure if I understand what you're saying here keith. 

4/3 x pi x (h/2)^3  = pi x h^3 / 6

So your initial answer was definitely correct.


----------



## bellenuit (14 June 2015)

cynic said:


> I'm not sure if I understand what you're saying here keith.
> 
> 4/3 x pi x (h/2)^3  = pi x h^3 / 6
> 
> So your initial answer was definitely correct.




Yes, I'm confused too. Both formulae are the same


----------



## keithj (16 June 2015)

cynic said:


> I'm not sure if I understand what you're saying here keith.
> 
> 4/3 x pi x (h/2)^3  = pi x h^3 / 6



Sorry - meant to type 4/3 x pi x *r* ^ 3. In my 1st example r did equal h/2.


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## cynic (19 June 2015)

E & 9. The other two cannot possibly violate the stated rule.


----------



## bellenuit (19 June 2015)

> TAKE a look at the two wheels in the image above. Which one is moving faster?




As it could be an optical illusion, the easiest way to check is to look at the image type. It is a JPG, so the images aren't moving at all. It would need to be a GIF or some other non-static image type to actually have the wheels moving.


----------



## bellenuit (19 June 2015)

cynic said:


> E & 9. The other two cannot possibly violate the stated rule.




Correct.


----------



## trainspotter (19 June 2015)

bellenuit said:


> As it could be an optical illusion, the easiest way to check is to look at the image type. It is a JPG, so the images aren't moving at all. It would need to be a GIF or some other non-static image type to actually have the wheels moving.




It is an optical illusion ... most people look at "A" and notice that "B" is rotating and vice versa. Peripheral vision causes this phenomenon. It is widely accepted by most psychologists that if both wheels are spinning together you are clinically insane.


----------



## cynic (20 June 2015)

Firstly treat the board as a 33 X 33 groups of 3 X 3. (i.e. 1089 groups of 9 squares)

(i) If there are no non-empty groups place a checker in the last available square of a group, otherwise place the checker in the central square of an empty 3 X 3 group. A further group has now been eliminated from play leaving an even number of playable groups.

Whenever the opponent responds by placing a checker on a central square within another empty group, repeat step (i).

(ii) Whenever the opponent places a checker on anything other than the central square within one of the vacant 3 X 3 groups, place a checker a chess knight's jump away ("L" shape) to the opponent's checker within that same group. This will leave exactly one more play within that group ensuring that an even number of groups remain in play. 

As there are 1089 groups of 3 X 3, strict subscription to the aforementioned rules will result in the commencing player always having the last available move.


----------



## keithj (20 June 2015)

I treated the board as two identical halves - either 2 triangles or 2 rectangles with a line of squares along the axis.

i) choose a line of symmetry (either horz, vert or a diagonal)
ii) place the 1st checker in the centre on the board.
iii) whenever the opponent places a checker NOT on your chosen line of symmetry, you place yours in the symmetrically opposite position.
iv) whenever the opponent places a checker on the line of symmetry,  you place yours also on the line of symmetry, but the same distance away from the centre as the opponents.

TL;DR;
The board must always have reflective symmetry.  After taking up the only square (the centre) that cannot copied, you simply copy the opponent in the opposite half of the board.

I guess you could also use rotational symmetry.


----------



## bellenuit (20 June 2015)

keithj said:


> I treated the board as two identical halves - either 2 triangles or 2 rectangles with a line of squares along the axis.
> 
> i) choose a line of symmetry (either horz, vert or a diagonal)
> ii) place the 1st checker in the centre on the board.
> ...




I haven't worked through Cynic's detailed answer yet, but Keith's answer is the one I know that will ensure a win.

You must place the first draught (should I be calling them checkers?) in the centre square and then you place all remaining draughts symmetrically opposite what your opponent played. I worked on the basis of diagonally opposite (e.g. if your opponent placed a dart 6 up and 3 across to the right from the centre piece, you place yours 6 down and 3 to the left from the centre piece). You could also, as Keith suggested, place the draughts horizontally or vertically opposite, with the adjustment he suggested for draughts placed on the line of symmetry. Doing it diagonally opposite does that automatically without the need to state that special case.

Once you place the centre draught, you are effectively then imitating on one side of the board what your opponent has done on the other side. And since after you place your piece, one side of the board will be a reflection of the other, then by definition, if your opponent is able to place a piece on the board that abides by the rules, you will be able to do so as well as the same symmetrically opposite position will also be available to you. Your opponent will always be the first to encounter a situation where he cannot place a piece, so you will always win.


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## cynic (20 June 2015)

bellenuit said:


> I haven't worked through Cynic's detailed answer yet, but Keith's answer is the one I know that will ensure a win.
> ...



I've just noticed a flaw in my method which would definitely require additional rules (adding to its complexity) in order to ensure success.

Keith's answer is ,of course, a far less complex and more sound approach to this problem. 
(Congratulations again Keith!)


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## cynic (23 June 2015)

They chose the numbers 6 & 7.

On day 1 (when nobody leaves) they realise that neither could have chosen 11-13 because they are all greater than 10 meaning that the other could be determined by subtraction from 13.

After day 2 they know that nobody could have chosen 0-2 as these can only contribute to a sum of 10 now that 11-13 have been eliminated.

After day 3 they know that nobody has chosen 8-10 as this can now only contribute to the sum of 13 since 0-2 have been eliminated.

After day 4 they know that  nobody has chosen 3-5 because these can now only contribute to a sum of 10 since 8-10 have been eliminated.

This leaves only the combination 6 & 7.


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## bellenuit (24 June 2015)

cynic said:


> They chose the numbers 6 & 7.
> 
> On day 1 (when nobody leaves) they realise that neither could have chosen 11-13 because they are all greater than 10 meaning that the other could be determined by subtraction from 13.
> 
> ...




Congratulations! Perfect answer.


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## cynic (27 June 2015)

2,3,4 & 5.

There cannot be more than 2 yellow marbles without the sum exceeding 17. So it is known that there are either 1 or 2 yellow marbles. 

The fact that Mary asked whether there was more than 1 yellow indicates that Mary has noticed that the house number can be produced from valid configurations containing either number of yellow marbles. 

Therefore the house number is 120 as this is the only product that occurs for valid configurations containing exactly 1 yellow and also for a valid configuration containing exactly 2.

The fact that Mary instantly knew from the Father's reply indicates that the number of yellow marbles must be 2 (had there been 1 then Mary would have needed further information in order to to decide between the 2 valid configurations containing 1 yellow that produce the number 120).

The only valid configuration for 2 yellow, which has a product corresponding to 120 and a sum less than 18, is the one  containing 3 blue, 4 green and 5 red marbles.


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## keithj (27 June 2015)

cynic said:


> 2,3,4 & 5.



Congrats.  I agree.  The additional bit of info that Mary requires (1 or 2 yellows) is sufficient to distinguish between the duplicate 120s. All the others products are unique & don't require the additional info.


```
Y	B	G	R	Product	Sum
1	2	3	4	24	10
1	2	3	5	30	11
1	2	3	6	36	12
1	2	3	7	42	13
1	2	3	8	48	14
.........
1	3	4	6	72	14
1	3	4	7	84	15
1	3	4	8	96	16
1	3	4	9	108	17
1	3	5	6	90	15
1	3	5	7	105	16
[B]1	3	5	8	120	17
1	4	5	6	120	16[/B]
1	4	5	7	140	17
[B]2	3	4	5	120	14[/B]
2	3	4	6	144	15
2	3	4	7	168	16
2	3	4	8	192	17
2	4	5	6	240	17
```


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## bellenuit (27 June 2015)

cynic said:


> 2,3,4 & 5.
> 
> There cannot be more than 2 yellow marbles without the sum exceeding 17. So it is known that there are either 1 or 2 yellow marbles.
> 
> ...




Damn. I thought this one would survive the weekend, but only lasted overnight. Congratulations again Cynic on being first and Keithj on arriving at the same answer.


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## luutzu (28 July 2015)

An easy one that can be solved with no math, just some logic.

A plane flies in a straight line from Perth to Sydney, then back in a straight line from Sydney to Perth. It travels with a constant engine speed (constant air speed) and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from Perth to Sydney?

----

Since air speed is constant, should be the same either way.

But if engine speed being constant mean the engine revs x per minute and that will speed the plane at say 800km/hr, then will be slower from Sydney as the headwind will slow it down. But assuming that engine will rev up/down to keep a constant air speed (it goes at 800km/h regardless), then same time.

But then if Bronwyn is at the gate and there's no front seat in first class, Perth to Sydney will be delayed a bit.


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## keithj (28 July 2015)

bellenuit said:


> *Increasing the Ratio of Women in the Population.
> 
> A king (dirty old leech) decided that there should be proportionally more women in his kingdom than the current 50%. He knows that the probability of a girl at birth is exactly 50% and that males and females have on average exactly the same lifespan (in his kingdom anyway), so he assumes this strategy will work.
> 
> ...



Q1.  His strategy will make no difference - the ratio will stay the same at 50%.  Each row below represents one set of parents, and each column their successive offspring (stopping when a Girl is produced).  The ratio of total number of offspring remains at 50%

B	B	B	B
B	B	B	G
B	B	G	
B	B	G	
B	G		
B	G		
B	G		
B	G		
G			
G			
G			
G			
G			
G			
G			
G			

Count of B = 15
Count of G = 15


Q2. still thinking...


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## bellenuit (28 July 2015)

luutzu said:


> An easy one that can be solved with no math, just some logic.
> 
> A plane flies in a straight line from Perth to Sydney, then back in a straight line from Sydney to Perth. It travels with a constant engine speed (constant air speed) and there is no wind. Will its travel time for the same round trip be greater, less, or the same if, throughout both flights, at the same engine speed, a constant wind blows from Perth to Sydney?
> 
> ...




So that we are all on the same wavelength, a constant engine speed means a constant air speed, that is the speed of the plane relative to the air it is passing through. Ground speed is the speed of the plane relative to the ground, which will be the air speed + or - the wind speed, depending on whether there is a tail or head wind.


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## luutzu (28 July 2015)

bellenuit said:


> So that we are all on the same wavelength, a constant engine speed means a constant air speed, that is the speed of the plane relative to the air it is passing through. Ground speed is the speed of the plane relative to the ground, which will be the air speed + or - the wind speed, depending on whether there is a tail or head wind.




Then going back to Perth will be slower as there's headwind.

That both way need to cover the same ground, going to Sydney with the wind meant no headwind (wind don't speed up the plane from behind, i don't think) while going back to Perth faces a drag so ground speed will be slower so will take longer over same distant.


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## cynic (28 July 2015)

I would have thought that the round trip would be theoretically the same. The speed lost flying into the headwind would be compensated by the speed gained on the reverse journey.


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## cynic (28 July 2015)

keithj said:


> Q1.  His strategy will make no difference - the ratio will stay the same at 50%.
> ...
> 
> Q2. still thinking...



I agree that the ratio will remain approximately constant throughout the generations. A further downside to the edict is that it caps the female population which would likely cause a slow decline in the total population as every so often a couple will fail to produce a girl.

As to Q2. I do not know an optimal solution but believe that if each couple were to reproduce until they have one less boy than girls, then the proportion of women in the populace would gradually increase.


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## bellenuit (29 July 2015)

cynic said:


> I would have thought that the round trip would be theoretically the same. The speed lost flying into the headwind would be compensated by the speed gained on the reverse journey.




Nope. It will always take longer if there is a constant wind speed in one direction over the complete trip than if there is no wind speed. You could take some theoretical air speeds and wind speeds and work out, but the reason I said that no math is needed is to use values that require no calculation (or minimal) and see where that leads.

If the Air Speed is A and the Wind Speed is W and the distance between Perth and Sydney is D, then in the case where there is no wind, the time to complete the return journey is 2D/A. In the second example, where there is a constant wind speed between Perth and Sydney, the total time is D/(A+W) + D/(A-W). Note that the divisor of all three elements (A, A+W, A-W) is the ground speed of the plane.

As I said you can work out for various combinations of A and W, but it should be obvious if you take some limiting examples logic will override the need for complex computation. Just imagine if the Wind speed were the same as the Air speed of the plane. Obviously the outbound trip to Sydney would only take half the time as previously as your ground speed is double what it was when there was no wind. But for the return journey, your ground speed is effectively zero (A-W) where A=W. So technically, once airborne, the plane stays still relative to the ground and it would take an infinite time to get back from Sydney to Perth.

Getting a bit more mathematical, it also should be obvious that the two elements to compute the time in the second return trip change at different rates, so that their sum is not constant.

When the Wind speed is zero, the out bound and return times are the same: D/(A+W) = D/(A-W) = D/A (as W=0)

But as the Wind speed increases from zero to the Air speed, the outbound time DECREASES by a limited amount, from D/A to D/(A+W)= D/2A. But the return journey time INCREASES at a vastly higher rate, from D/A to D/(A-W) which is infinity when A=W. So if it took 5 hours when no wind, then as the wind speed increases from 0 to the Air speed, the outbound time decreases from 5 to 2.5 hours and the return increases from 5 to infinity.

If you want practical proof, the next time you are at an airport with a long travelator, you can time yourself walking beside the travelator but not on it and then walking on it in the same direction and on it in the opposite direction. If you can maintain a constant walking pace and are not arrested, the sum of the latter two will be more than twice the former. In this case the travelator represents the wind speed and your walking pace is the air speed. It is also obvious if your pace is the same as that of the travelator (A=W), when walking against it, you get nowhere.


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## bellenuit (29 July 2015)

cynic said:


> I agree that the ratio will remain approximately constant throughout the generations.






> A further downside to the edict is that it caps the female population which would likely cause a slow decline in the total population as every so often a couple will fail to produce a girl.




I find these sentences contradictory. If the probability of a boy (or girl) is exactly 50%, then there will always be the same amount of boys and girls as keithj correctly showed. 



> As to Q2. I do not know an optimal solution but believe that if each couple were to reproduce until they have one less boy than girls, then the proportion of women in the populace would gradually increase.




Again not possible as keithj's example demonstrates. The answer to Q2 is that there is NO edict of the type given in Q1 that can change the ratio. If we assume for simplicity that each couple can produce one child per year, then if we begin with an equal number of males and females and the probability of a boy (or girl) at birth is exactly 50%, then after the first year 50% of new babies will be boys and 50% girls. This means the overall ratio in the population is still 50%, so at the beginning of year 2 we are exactly where we were at the beginning of year 1. So there is no edict you can give at the start of year 2 (which is what your Q2 solution implied) that could not have been given at the start of year 1.


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## cynic (29 July 2015)

I'm okay with your windspeed explanation, but I think I'm viewing things from a slightly different angle on the other problem. 

If a couple produces an odd number of children there will always be an imbalance in the numbers of boys and girls. All the edict has to do is demand that each couple reproduce until that imbalance favours girls in their particular family and then stop. Or perhaps I'm misunderstanding something!?


----------



## bellenuit (29 July 2015)

cynic said:


> I'm okay with your windspeed explanation, but I think I'm viewing things from a slightly different angle on the other problem.
> 
> If a couple produces an odd number of children there will always be an imbalance in the numbers of boys and girls. All the edict has to do is demand that each couple reproduce until that imbalance favours girls in their particular family and then stop. Or perhaps I'm misunderstanding something!?




Cynic, if you list possible outcomes of babies born like keithj did, you would see that is is not possible to produce the imbalance you suggest with an exact 50% probability of a girl.

Think of what I said in my last response. At the end of each year we are back to square one, so there is no edict to use in year 2 that couldn't have been used in year one.

Your example. Starting point: #Males = #Females. At end of first year, all couples will have one baby, half of which are boys. They all now have an odd number of children and 50% of those will be girls. But #Males still = #Females. Those with girls stop producing (your solution) and remainder (those who had boys) have a baby the second year.  Half of these will again be girls (so #Males still = #Females), but as all these couples have an even number of children now, they reproduce once more. Again you will have 50% girls with no change to the ratio. Each will now have an odd number of children, so those with two girls and one boy are to stop reproducing. But we are still in exactly the same situation as we were at the start of year 1, when those with 1 girls and 0 boys were to stop reproducing. 

It's a bit like flipping a coin. The outcome is always 50% heads (or tails) and is never dependent on what came before. So if you have a thousand people flipping continuously, you can never change the probably outcome overall by telling some flippers that they have to stop flipping because of a particular sequence that they have experienced. At the end of the day, 50% of the flips, no matter who has done them should be heads and 50% tails.


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## cynic (29 July 2015)

19 minutes.

First A crosses with D taking 10 minutes.

A returns with flashlight in 1 minute then accompanies C across taking 5 minutes.

A again returns with flashlight in 1 minute then accompanies B across taking 2 minutes .

They are all now safely across taking a total of 10 + 1 +5 + 1 + 2 = 19 minutes.


----------



## barney (29 July 2015)

cynic said:


> 19 minutes.
> 
> First A crosses with D taking 10 minutes.
> 
> ...





That's what I thought unless there is some sneaky formula

My first thought however was .... A) is obviously fit if he can do it in 1 minute ..... He could just piggy back the other three guys across = 5 minutes + a bit of extra time to accommodate for exhaustion


----------



## keithj (29 July 2015)

barney said:


> That's what I thought unless there is some sneaky formula
> 
> My first thought however was .... A) is obviously fit if he can do it in 1 minute ..... He could just piggy back the other three guys across = 5 minutes + a bit of extra time to accommodate for exhaustion



A & B cross taking 2 mins
A returns in 1 min
C & D go across in 10 mins
B returns in 2 mins
A & B go across in 2 mins

Total of 17 mins.

This minimises the return times, and also minimises the time taken for the slowest 2.


----------



## bellenuit (29 July 2015)

keithj said:


> A & B cross taking 2 mins
> A returns in 1 min
> C & D go across in 10 mins
> B returns in 2 mins
> ...




Correct keithj.

Alternatively you could swap when A and B returns and also get the same answer of 17 minutes.


----------



## cynic (29 July 2015)

What's a couple of minutes between friends?!

Congratulations again Keith.


----------



## barney (29 July 2015)

keithj said:


> A & B cross taking 2 mins
> A returns in 1 min
> C & D go across in 10 mins
> B returns in 2 mins
> ...




Ahh ..... very good!!


----------



## cynic (29 July 2015)

3 links of one of the 3 link lengths need to be opened. Each link can then be used to join the remaining 3 chain lengths into a circle.


----------



## bellenuit (29 July 2015)

cynic said:


> 3 links of one of the 3 link lengths need to be opened. Each link can then be used to join the remaining 3 chain lengths into a circle.




Correct.


----------



## cynic (3 August 2015)

Q1.  5 cms


----------



## cynic (3 August 2015)

cynic said:


> Q1.  5 cms



Q2. 20 sq cms


----------



## cynic (3 August 2015)

cynic said:


> Q2. 20 sq cms




Q3. 10 cms


----------



## pixel (3 August 2015)

cynic said:


> Q3. 10 cms




Q4 = 37 cm^2

Q1, 2, 3 agreed


----------



## pixel (3 August 2015)

pixel said:


> Q4 = 37 cm^2
> 
> Q1, 2, 3 agreed




Q5: 35 cm^2


----------



## cynic (3 August 2015)

pixel said:


> Q5: 35 cm^2




+ 1 and congratulations on beating me to the tougher ones.


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## bellenuit (3 August 2015)

pixel said:


> Q5: 35 cm^2




For those who are having difficulty working out the solution without resorting to computing interim fractions or using algebra, this is how Number 3 is worked out. 





You can easily calculate the missing rectangle in the top left hand corner (delineated in red) as its sides are obviously 1 X 5. It has a width 1 because its width is the difference between the widths of the rectangle immediately to its right (4) and that below it (5). It has a height of 5 because it has the same height as the rectangle beside it, which must be 20 divided by 4. 

If you then add the areas of this new rectangle with those of the rectangle to its left and the rectangle below it, you get a new combination rectangle of 5 + 20 + 14 or 39cm2. This combination rectangle is exactly half the area of the large rectangle on the right (78cms) and as it shares the same height, its width then must be half that of the bigger rectangle. But we know the combination rectangle has a width of 5, therefore the bigger one must have a width of 10cm, which is the answer we are looking for.

According to the source article, the reason these puzzles are to be solved without use of fractions or algebra, basically with just the knowledge that the area of a rectangle equals its width by height, is that these are problems set for Japanese school kids who haven't yet got to the stage where they have learned about algebra or fractions.


----------



## skc (3 August 2015)

pixel said:


> Q4 = 37 cm^2




Got everything else except this one. A hint please??

(Hopefully I can solve it before a response!)



bellenuit said:


> According to the source article, the reason these puzzles are to be solved without use of fractions or algebra, basically with just the knowledge that the area of a rectangle equals its width by height, is that these are problems set for Japanese school kids who haven't yet got to the stage where they have learned about algebra or fractions.




Thanks Bellenuit. I enjoyed them. It was a good workout for the brain.


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## cynic (4 August 2015)

I used formulae for #4.

L1 = ?
L2 = 4

W1 = ?
W2 = ?
W3 = ?

W2 + w3 = 10
4 * (w1 + w2) = 29
L1 * w3 = 43
L1 * w1 = 21

L1 * (w3-w1) = 22

Etc


----------



## skc (4 August 2015)

cynic said:


> I used formulae for #4.




I worked it out with algebra too. But supposedly you can work it out without such... Anyone else?


----------



## bellenuit (4 August 2015)

skc said:


> I worked it out with algebra too. But supposedly you can work it out without such... Anyone else?




Official answer to Number 4 ....




_The left rectangle with a blue border covers the rectangle with area 21 and a section of the rectangle below. Imagine placing an identical rectangle to the right of the grey rectangle. The shaded blue area must have area 21, and the dotted blue area must be z.

The area x is 43 - 21 = 22

The area of the rectangle with the red border is 4 x 10 = 40. But it is also 29 + y, since a + z = 29. Therefore y = 11

So x is equal to double y. Which means that any rectangle above the horizontal line must have twice the area of the rectangle beneath the line that it shares a side with.

The green rectangle must have area double the rectangle underneath it, which has area 29. So the green rectangle has area 2 x 29 = 58

The missing value is 58 - 21 = 37cm2_

An answer hasn't been provided for Number 5, other than saying the readers should work it out for themselves. I suspect it is too complex to explain the solution by a non-algebraic method, so have left it as a teaser for the readers to do who can then post their answers in the blog.


----------



## cynic (4 August 2015)

skc said:


> I worked it out with algebra too. But supposedly you can work it out without such... Anyone else?




Apologies.

The only thing that occurs to me outside of direct formulae is a brute force approach that presumes that the uppermost rectangles height are a whole multiple of the lower.


----------



## pixel (4 August 2015)

bellenuit said:


> For those who are having difficulty working out the solution without resorting to computing interim fractions or using algebra, this is how Number 3 is worked out.
> 
> View attachment 63718
> 
> ...




I didn't bother with the rectangle top left. For #3 we don't need a calculator:

The height of the 20sq has to be 20 / 4 = 5
The height of the 14sq has to be 14 / 5 = 2.8
add the two together, then divide 78 by the result --> 10


----------



## skc (4 August 2015)

bellenuit said:


> Official answer to Number 4 ....




Very nice. I tried to do something with the rectangle on the bottom right, but I didn't think of mirroring the thing...



bellenuit said:


> An answer hasn't been provided for Number 5, other than saying the readers should work it out for themselves. I suspect it is too complex to explain the solution by a non-algebraic method, so have left it as a teaser for the readers to do who can then post their answers in the blog.




#5 was actually easier than #4. It required you to "work around the board" to get to the dimensions of the rectangle in question... but it all involved just one method and didn't need any lateral thinking like #4.

It fact it was easy where, anytime you face a calculation involving fractions, you know you are on the wrong track.


----------



## cynic (4 August 2015)

Given that algebra is essentially a methodology for expressing and determining unknown quantities from known, I find it amusing that all solutions thus far are essentially based upon the same underlying problem solving principle, irrespective of whether they were algebraically expressed.


----------



## bellenuit (4 August 2015)

pixel said:


> I didn't bother with the rectangle top left. For #3 we don't need a calculator:
> 
> The height of the 20sq has to be 20 / 4 = 5
> The height of the 14sq has to be 14 / 5 = 2.8
> add the two together, then divide 78 by the result --> 10




Yes, but remember the challenge was to solve the problem without using fractions (2.8 being an implied fraction). I didn't explicitly state that as a rule, as I wasn't quite sure if that was what was intended, but I did add...

For Number 1 I stated....

_Note: For all puzzles, all answers are whole numbers (no fractions). The instructions are a bit ambiguous, but I also understand that to solve the puzzles you do not need to deal with fractions.
_
And for Number 3 I added...

_Note: I can confirm that having now solved Number 3, that 1, 2 and 3 can be solved without resorting to using fractions in interim results. I assume the same applies to 4 and 5._

Nevertheless, congrats on your correct answers.


----------



## keithj (28 August 2015)

bellenuit said:


> A fairly easy one.
> 
> View attachment 63788
> 
> ...



NO.

Assume the board is a chess board with alternating black & white squares.  A domino must cover one black square & one white square. 

The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.


----------



## cynic (28 August 2015)

keithj said:


> NO.
> 
> Assume the board is a chess board with alternating black & white squares.  A domino must cover one black square & one white square.
> 
> The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.




I did very much suspect that No was the answer but couldn't figure out a way to prove or disprove (apart from quasi brute force).


----------



## bellenuit (28 August 2015)

keithj said:


> NO.
> 
> Assume the board is a chess board with alternating black & white squares.  A domino must cover one black square & one white square.
> 
> The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.




Yep, good on you, that is the smart way to solve it. If the squares are alternate colours like a chess board, then diagonally opposite corners are the same colour so long as the number of rows and columns are even, which is the case here. So depending on whether the corner squares removed are white or black, there will be 2 less of that colour than the other. And as you said, since a domino always cover two squares of opposite colours, then each time a domino is placed, there will always be the same number of each colour now covered. So the task is impossible if there are not the same number of squares of each colour.


----------



## skc (28 August 2015)

keithj said:


> NO.
> 
> Assume the board is a chess board with alternating black & white squares.  A domino must cover one black square & one white square.
> 
> The board has 400 squares - 200 black & 200 white. Two black ones are already covered. There are only 198 vacant black ones, but 200 white ones. Therefore there is NO arrangement of dominos that can cover the whole board.




Thanks. I like the simplicity of the solution.


----------



## cynic (31 August 2015)

Correction to my previous answer. PI/16 is the probability that the coin doesn't overlap adjacent squares. Needless to say 1-(PI/16) is the probability that it does.


----------



## cynic (31 August 2015)

cynic said:


> Correction to my previous answer. PI/16 is the probability that the coin doesn't overlap adjacent squares. Needless to say 1-(PI/16) is the probability that it does.




Reasoning is that the centre of the coin will land in a precise place within or on the boundaries of one square. There is circle of diameter of 1cm at the centre of every square within which the coins centre could land without the entire coin overlapping any adjacent squares.


----------



## cynic (31 August 2015)

cynic said:


> Correction to my previous answer. PI/16 is the probability that the coin doesn't overlap adjacent squares. Needless to say 1-(PI/16) is the probability that it does.






cynic said:


> Reasoning is that the centre of the coin will land in a precise place within or on the boundaries of one square. There is circle of diameter of 1cm at the centre of every square within which the coins centre could land without the entire coin overlapping any adjacent squares.




I just noticed that I probably should have used a 1cm square instead of a circle making the probabilities 1/4 and 3/4.


----------



## pixel (31 August 2015)

cynic said:


> I just noticed that I probably should have used a 1cm square instead of a circle making the probabilities 1/4 and 3/4.




I came to the same conclusion: one throw in four.
Rationale: The centre of the coin will need to be half the coin diameter inside either border. Assuming that "touching" the edges satisfies the condition, that makes the permitted locale for the coin centre exactly one square centimeter.


----------



## cynic (31 August 2015)

pixel said:


> I came to the same conclusion: one throw in four.
> Rationale: The centre of the coin will need to be half the coin diameter inside either border. Assuming that "touching" the edges satisfies the condition, that makes the permitted locale for the coin centre exactly one square centimeter.




Great pixels look alike, and cynics rarely differ.

Agree that there is a 3/4 probability that the coin will land straddling two or more squares.


----------



## bellenuit (31 August 2015)

Yes, 3/4 is correct for the reasoning given above. If the coin centre lands anywhere within the imaginary 1cm square that is central to each chessboard square then it will not straddle a border. Outside that it will. Since that square is 1/4 the size of the chessboard square, then overall 1/4 of the total board area is safe from straddling with 3/4 not. Congrats.


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## cynic (13 September 2015)

Jessica did it!

One cut from the side will cut the cake into two equal height cylinders, the next two cuts can be made in such a way that they slice both cylinders into quarters of equal size.


----------



## bellenuit (13 September 2015)

cynic said:


> Jessica did it!
> 
> One cut from the side will cut the cake into two equal height cylinders, the next two cuts can be made in such a way that they slice both cylinders into quarters of equal size.




Yes (only Nicole told the truth) and Yes. Congrats.


----------



## pixel (13 September 2015)

Q1: If only one girl told the truth, it has to be Anne, leaving Jessica the culprit.

Q2: A horizontal cut as #1, then two vertical at 90 degree angles will do the trick; but let's hope there's no icing on the cake because if there was, half the recipients - the ones getting the bottom pieces - will be disadvantaged.


----------



## bellenuit (13 September 2015)

pixel said:


> Q1: If only one girl told the truth, it has to be Anne, leaving Jessica the culprit.
> 
> Q2: A horizontal cut as #1, then two vertical at 90 degree angles will do the trick; but let's hope there's no icing on the cake because if there was, half the recipients - the ones getting the bottom pieces - will be disadvantaged.




Jessica is the culprit, but it isn't Anne who is telling the truth. 

*Anne yelled, "No, it was Sandy!"* How could you conclude Anne is telling the truth and at the same time say Jessica is the culprit? 

The only one telling the truth is Nicole and Jessica is the culprit.


----------



## pixel (13 September 2015)

bellenuit said:


> Jessica is the culprit, but it isn't Anne who is telling the truth.
> 
> *Anne yelled, "No, it was Sandy!"* How could you conclude Anne is telling the truth and at the same time say Jessica is the culprit?
> 
> The only one telling the truth is Nicole and Jessica is the culprit.




Sorry, I "remembered" the wrong name: Should've copied 







> *Nicole * (not Anne) said, "Sandy's a liar."



 to get the name right.


----------



## pixel (13 September 2015)

> No. 1
> 
> Anne and Jessica were working on the computer along with their friends Sandy and Nicole. Suddenly, I heard a crash and then lots of shouts. I rushed in to find out what was going on, finding the computer monitor on the ground, surrounded with broken glass!
> 
> ...




In that case, Anne is right, Sandy is wrong, which makes Jess and Nicole also right.
*Sandy dunnit.*


----------



## cynic (13 September 2015)

pixel said:


> In that case, Anne is right, Sandy is wrong, which makes Jess and Nicole also right.
> *Sandy dunnit.*




Yes, that was the answer that I arrived at.


----------



## skc (28 September 2015)

bellenuit said:


> This one requires a fair bit of thinking....
> 
> *Draw a side view of this wooden object
> 
> ...



*

Side view



*


----------



## bellenuit (28 September 2015)

skc said:


> Side view
> 
> View attachment 64497




Correct, but the triangle is a dotted line as it is not visible from the left side. I also made a mistake in the question when I said the square hole runs through the object, which obviously implies from one side to the other. I shouldn't have defined what it was other than say it was not some lines drawn on the object. It doesn't actually run through but is just cut into the object. Also, it only looks square when viewed from either the top or front. When viewed perpendicular to the sloping face it is rectangular. Apologies if that put anyone off.


----------



## skc (28 September 2015)

bellenuit said:


> Correct, but the triangle is a dotted line as it is not visible from the left side. I also made a mistake in the question when I said the square hole runs through the object, which obviously implies from one side to the other. I shouldn't have defined what it was other than say it was not some lines drawn on the object. It doesn't actually run through but is just cut into the object. Also, it only looks square when viewed from either the top or front. When viewed perpendicular to the sloping face it is rectangular. Apologies if that put anyone off.




Haha... I should have clarified that my little triangle meant to be dotted line. It's just that I couldn't find dotted line for shapes in the Microsoft Paint that I used to create the drawing, so I used a lighter line instead.


----------



## cynic (5 January 2016)

((10 Ã— 9 Ã— 8) - (7 Ã— 6) - 5 - 4 + 3 )Ã— (2 + 1)


----------



## cynic (5 January 2016)

cynic said:


> ((10 Ã— 9 Ã— 8) - (7 Ã— 6) - 5 - 4 + 3 )Ã— (2 + 1)




((10 - 9 +(8 Ã— 7 Ã— 6) -5 + 4) Ã— 3 Ã— 2)/1


----------



## cynic (5 January 2016)

cynic said:


> ((10 - 9 +(8 Ã— 7 Ã— 6) -5 + 4) Ã— 3 Ã— 2)/1




((10 Ã— 9 Ã— 8 Ã— 7 Ã— 6)/(5/(4 - 3)))/(2 + 1)


----------



## cynic (5 January 2016)

cynic said:


> ((10 Ã— 9 Ã— 8 Ã— 7 Ã— 6)/(5/(4 - 3)))/(2 + 1)




((10 Ã— 9 Ã— 8 Ã— 7 Ã— 6)/(5 Ã— (4 - 3)))/(2 + 1)


----------



## cynic (5 January 2016)

cynic said:


> ((10 - 9 +(8 Ã— 7 Ã— 6) -5 + 4) Ã— 3 Ã— 2)/1




Also Ã— 1 in place of /1


----------



## cynic (5 January 2016)

cynic said:


> Also Ã— 1 in place of /1




(10 - 9) Ã— (8 Ã— 7 Ã— 6) Ã— (5 - 4) Ã— 3 Ã— 2 Ã— 1

Also another 3 variations by substitution of / for Ã— with the (5 -4) and 1.


----------



## bellenuit (5 January 2016)

cynic said:


> (10 - 9) Ã— (8 Ã— 7 Ã— 6) Ã— (5 - 4) Ã— 3 Ã— 2 Ã— 1
> 
> Also another 3 variations by substitution of / for Ã— with the (5 -4) and 1.




Yep, this and all your previous variations verified = 2106 by Spotlight Search on my Mac. Off to a great start for 2016.


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## bellenuit (5 January 2016)

The person who posed the problem in the paper gave this as the most elegant solution he had seen:

(10 x 9 x 8 x 7 x 6)/(5 + 4 + 3 + 2 + 1) = 2016


----------



## cynic (5 January 2016)

bellenuit said:


> The person who posed the problem in the paper gave this as the most elegant solution he had seen:
> 
> (10 x 9 x 8 x 7 x 6)/(5 + 4 + 3 + 2 + 1) = 2016




Much as I  would love to have found all of em, I  expected there were likely to be many other solutions, however, I  am genuinely surprised that one escaped my attention.

I  agree with the writer's comments on elegance.


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## cynic (13 January 2016)

The answer is yes. Anna must either be married or unmarried either way the answer will become yes.


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## bellenuit (13 January 2016)

cynic said:


> The answer is yes. Anna must either be married or unmarried either way the answer will become yes.




Correct. I was one of the 80% that got it wrong. I was too quick to jump to conclusions.


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## cynic (13 January 2016)

bellenuit said:


> Correct. I was one of the 80% that got it wrong. I was too quick to jump to conclusions.




I  am guessing that many would have opted for unable to determine. That was my first inclination before examining it a bit more thoroughly.


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## pixel (25 January 2016)

Try it this way:

*A dozen, a gross, and a score
increased by three square roots of four,
divided by seven,
plus five times eleven
will give you nine squared, nothing more.*

PS: It's true  182 / 7 = 26; add 55; result is 81


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## bellenuit (25 January 2016)

pixel said:


> Try it this way:
> 
> *A dozen, a gross, and a score
> increased by three square roots of four,
> ...




Yep, great effort and almost identical to the one I read in the paper that posed the problem (see below).





_A dozen a gross and a score
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is equal to nine squared, and no more._


----------



## pixel (26 January 2016)

bellenuit said:


> Yep, great effort and almost identical to the one I read in the paper that posed the problem (see below).
> 
> View attachment 65638
> 
> ...




well, the similarity isn't surprising.
Once you have the "names" of the first 3 numbers lined up, the rest falls more or less in place.

... it probably helps that I've dabbled in composing and translating Limericks for fun.
couple examles:


> An investor, betrayed by his broker,
> cracked his skull with a brass fire poker.
> His defense was "This clown
> just kept averaging down.
> ...


----------



## cynic (26 January 2016)

Well, I  am now glad that I didn't complete my answer to that one. What I had forming was somewhat creative but terribly clunky by comparison.

Well done pixel!


----------



## keithj (1 February 2016)

SirRumpole said:


> I'm sure the people here will solve this in an instant !
> 
> 'No winning answers yet' to British spy agency's Christmas puzzle challenge
> 
> http://www.abc.net.au/news/2016-01-...o-british-spy-agency-christmas-puzzle/7127758



Complete puzzle here http://www.gchq.gov.uk/press_and_media/news_and_features/Pages/Directors-Christmas-puzzle-2015.aspx


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## skc (3 February 2016)

bellenuit said:


> A shadowy puzzle
> 
> *You have a light source that throws light out in all directions. You are to construct a closed room around that light source using straight perpendicular walls which meet each other at corners. The shape you end up with must be such that every wall is either partly or fully shaded from the light source (e.g.  every wall of the room is either partly or fully shaded from the light source by other walls in the room).
> 
> ...




Something like this? Not drawn perfectly but the idea is to have a deep enough nook at each corner that will cast a shadow on each wall.


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## bellenuit (4 February 2016)

skc said:


> Something like this? Not drawn perfectly but the idea is to have a deep enough nook at each corner that will cast a shadow on each wall.
> 
> View attachment 65765




That is a correct answer, but can be done with less walls. I hope I haven't confused people by describing the walls as perpendicular. I meant perpendicular to the ground, not perpendicular to each other. They do not have to meet at 90 degree angles. The reason I specified perpendicular (to the ground) was that some people who were attempting to solve the problem when it was posed in the paper I read had asked if the walls could be at other than 90 degrees to the ground and such solutions were excluded by the problem poser. I will update the problem with this info.


----------



## keithj (4 February 2016)

bellenuit said:


> That is a correct answer, but can be done with less walls.



I can't see it being done with less than 6 walls.


----------



## bellenuit (4 February 2016)

keithj said:


> I can't see it being done with less than 6 walls.
> 
> 
> 
> ...




The one on the right is what I also came up with and many who replied to the newspaper column had the same or variations of it. That has 6 sides. The poser claims that it can be done with 5 walls, but hasn't produced a 5 walled solution yet. If he does I will post it here.


----------



## skc (4 February 2016)

bellenuit said:


> That is a correct answer, but can be done with less walls. I hope I haven't confused people by describing the walls as perpendicular.




That was how I interpreted the question.. Doesn't really change much and still a fun exercise nonetheless.



bellenuit said:


> The one on the right is what I also came up with and many who replied to the newspaper column had the same or variations of it. That has 6 sides. The poser claims that it can be done with 5 walls, but hasn't produced a 5 walled solution yet. If he does I will post it here.




Might give this a shot when I have time.


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## keithj (4 February 2016)

bellenuit said:


> The poser claims that it can be done with 5 walls, but hasn't produced a 5 walled solution yet.



Logically, the best possible solution.....

...can only have walls of 2 types -

A wall must either shade another wall AND be in complete shade itself
or a wall must be partially shaded by another wall.
In an optimal solution, a wall can only shade a max of 1 other wall. (Otherwise the 2 walls it shades can be straightened to make a single partially shaded wall)
In an optimal solution, 2 adjacent completely shaded walls can be straightened to into 1 wall.

Therefore the wall types must alternate....

...therefore there must be an even number of walls.
So I'd be impressed to see a 5 walled solution.


----------



## bellenuit (4 February 2016)

keithj said:


> Logically, the best possible solution.....
> 
> ...can only have walls of 2 types -
> 
> ...




You are correct. The minimum is 6. The original poser tweeted "A prize for whoever can do it with 5 walls". Since prizes are never offered for his puzzle answers, it seems that was made jokingly, but obviously p***ed off those who took it seriously (I being one, as I have wasted a lot of time on a 5 solution subsequent to working out the 6 wall solution). He has some interesting puzzles but is terrible in formulating them, usually only getting it right after several readers point out obvious mistakes in his puzzles.


----------



## cynic (14 March 2016)




----------



## bellenuit (14 March 2016)

cynic said:


> View attachment 66065




Well done! That's what I got too. 

Cynic, do you do Kakuro? I do Sudoku daily and can usually get the most difficult within an hour (not the above one, which took me about 4 hours). However, with Kakuro, I struggle on the simplest categories, rarely finishing what The West Australian categorise as Gentle or Moderate. They also have Tough and Diabolical, which I don't even attempt.

I have read all the tips and techniques that I have found on the internet, but still have huge difficulty solving them.

How do you fare on Kakuro if you do them, particularly in comparison to Sudoku?


----------



## cynic (14 March 2016)

bellenuit said:


> Well done! That's what I got too.
> 
> Cynic, do you do Kakuro? I do Sudoku daily and can usually get the most difficult within an hour (not the above one, which took me about 4 hours). However, with Kakuro, I struggle on the simplest categories, rarely finishing what The West Australian categorise as Gentle or Moderate. They also have Tough and Diabolical, which I don't even attempt.
> 
> ...





This particular puzzle took me a similar amount of time to yourself. I  somehow kept making the same mistake part way into it and having to start afresh. I  was almost comvinced that there was an error with the actual puzzle. Luckily I persevered with it and finally got past the glitch in my workings.

I  have only done a small handful of kakuro over the years, but, the one's I've encountered weren't particularly difficult and would probably fall into the gentle category.

Edit: forgot to mention that by comparison I do find kakuro easier than the harder classes of sudoko problems.


----------



## cynic (15 March 2016)

The larger quarter circle has double the radius of each of the two half circles. So the total surface are of the quarter circle works out to (pi x (2r)^2)/4) = pi x (r^2). Each of the two identical half circles works out to (pi x (r^2))/2. 

So the two half circles would add up to the same area as the quarter if only they didn't overlap. The area taken up by these semicircles is equal to 2 x pi x (r^2) - (area of overlap). Hence the residual of the quarter circle has an area equal to (pi x (r^2)) - (((2 x pi x (r^2)/2) - (area of overlap)) = (area of overlap). Ie. The blue area must equal the red area.


----------



## bellenuit (15 March 2016)

cynic said:


> The larger quarter circle has double the radius of each of the two half circles. So the total surface are of the quarter circle works out to (pi x (2r)^2)/4) = pi x (r^2). Each of the two identical half circles works out to (pi x (r^2))/2.
> 
> So the two half circles would add up to the same area as the quarter if only they didn't overlap. The area taken up by these semicircles is equal to 2 x pi x (r^2) - (area of overlap). Hence the residual of the quarter circle has an area equal to (pi x (r^2)) - (((2 x pi x (r^2)/2) - (area of overlap)) = (area of overlap). Ie. The blue area must equal the red area.




Yes, that turned out to be quite easy. My method was pretty much the same as yours.


----------



## cynic (6 June 2016)

1 to D1
T to B4
2 to A1
3 to B1
T to D4


----------



## bellenuit (6 June 2016)

cynic said:


> 1 to D1
> T to B4
> 2 to A1
> 3 to B1
> T to D4




Yep. Well done.


----------



## cynic (6 June 2016)

bellenuit said:


> Yep. Well done.




Correct me if mistaken, but there appears to be several solutions to this one.

Move 3 could be 2 to A2 or A3

Then move 4 could be 3 to A1 or A2


----------



## bellenuit (6 June 2016)

cynic said:


> Correct me if mistaken, but there appears to be several solutions to this one.
> 
> Move 3 could be 2 to A2 or A3
> 
> Then move 4 could be 3 to A1 or A2




Yes, there is. Moves 1 and 2 are key. Then its just a matter of getting pieces 2 and 3 anywhere out of the way so that T can be slid down.


----------



## pixel (7 June 2016)

1 to D1
T to the right
2 to A3
3 to A2
T down into the corner


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## cynic (20 June 2016)

(128-1) = 127

12


----------



## pixel (20 June 2016)

Wimbledon
Q1: 127 = 64+32+16+8+4+2+1, or 128-1
Q2: 24 = 4 aces or 4 lethal first returns per game * 6 games per set


----------



## pixel (20 June 2016)

pixel said:


> Wimbledon
> Q1: 127 = 64+32+16+8+4+2+1, or 128-1
> Q2: 24 = 4 aces or 4 lethal first returns per game * 6 games per set




Correction: Q2 needs only 12 if the opponent double-faults his/her 12 serves 

hat-tip to cynic


----------



## bellenuit (20 June 2016)

Cynic takes the winnings and the extra points..

You are both correct on the first. 127 matches. Because it is a knockout, each participant must lose 1 match, except the final winner. Hence 127 matches lost or as Cynic put it (128-1). That was the easier way to solve this problem. 

12 is the answer to part 2. Although it takes you a minimum of 4 hits to win a game, you only need to hit the ball in the 3 games you serve, thus 12 hits. Your opponent can double fault every point of the 3 games he serves without you needing to hit the ball.

PS. Pixel, our posts crossed


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## bellenuit (21 June 2016)

Retraction. The answer to part 2 is once. You only need to hit the ball once to win a set. This is the explanation (BTW, I thought 12 too):

_Imagine if every time you serve you swing at the ball and miss. This is a fault. Do this twice in a row and you lose the point. Do it eight times in a row and you lose the game.

Now imagine you do this for your six service games and that your opponent also double faults on every serve. The score after 12 games will be 6-6 and the set will enter the tiebreak.

If you continue to swing at the ball and miss, and your opponent double faults, neither of you will be able to win the tie-break. But depending on who serves first you will either get to 6-5, or 7-6 ahead. When you you do, ace that serve and you will have won the set with your racket hitting the ball only once._


----------



## pixel (21 June 2016)

bellenuit said:


> Retraction. The answer to part 2 is once. You only need to hit the ball once to win a set. This is the explanation (BTW, I thought 12 too):
> 
> _Imagine if every time you serve you swing at the ball and miss. This is a fault. Do this twice in a row and you lose the point. Do it eight times in a row and you lose the game.
> 
> ...




Oh Em Gee :1zhelp: That would be some match to watch (not!)
Would either player have made it to Wimbledon though?


----------



## cynic (21 June 2016)

Well it was a nice illusion whilst it lasted, but then I never was very good at tennis.


----------



## bellenuit (21 June 2016)

pixel said:


> Oh Em Gee :1zhelp: That would be some match to watch (not!)




On a par with listening to some of the speeches given this election.


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## cynic (1 August 2016)

One prisoner always predicts the opposite of his flip. The other prisoner always predicts the exact result of his flip. They both continue to live.


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## bellenuit (1 August 2016)

cynic said:


> One prisoner always predicts the opposite of his flip. The other prisoner always predicts the exact result of his flip. They both continue to live.




Correct.


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## cynic (13 September 2016)

8 cm

Forty third


----------



## bellenuit (13 September 2016)

cynic said:


> 8 cm
> 
> Forty third




8 cm - correct
Forty third - wrong


----------



## Habakkuk (13 September 2016)

bellenuit said:


> 8 cm - correct
> Forty third - wrong




Forty sixth


----------



## bellenuit (13 September 2016)

Habakkuk said:


> Forty sixth




Correct


----------



## cynic (13 September 2016)

bellenuit said:


> Correct




I stand corrected.

I neglected to count one of the the's. Hence the discrepancy.

Edit: congratulations habakkuk!


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## pixel (15 September 2016)

1) 8cm (2 covers plus the full width of volume II)
2: forty sixth (ignoring [...], the 2nd c is in "sentence", being the 36th letter, plus 10 letters for the number spelled in full)


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## cynic (10 October 2016)

(17+15+10)/2=21 games.

Every player will play at minimum one in every two games following the first.

"A" must have lost all 10 games beginning with the second game.

B & C played the first game.


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## bellenuit (10 October 2016)

cynic said:


> (17+15+10)/2=21 games.
> 
> Every player will play at minimum one in every two games following the first.
> 
> ...




Yes, you are right. Congratulations. Your logic is correct.

If A played in the first game, the minimum number of games he would play would be 11. This could only be arrived at by losing all games (1st, 3rd, 5th ... 21st) or winning just 1 game (and losing 10 of the remainder). Either of these implies a minimum of 11 games and any other possible outcome would be >11 games. The only way that he could have played just 10 games is to have *not* been in the first game. Then he would have been in the 2nd and lost all games he played (2nd, 4th, ... 20th). If he had won the second game, he would again play more than 10 games. 

A nice elegant solution as they say.


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## pixel (13 November 2016)

*We have to add one Red ball on the left and 8 Red balls (to make it 9) on the right.*

Calculation is simple:
4R + 2B = 2Y  =>  6Y = 6B + 12R
3R + 5B = 3Y  =>  6Y = 10B + 6R

subtraction gives
1Y = 3B - 1R  and  4B = 6R or 1B = 2/3 R

On the 3rd scale, we have 3 B + 1 Y, equivalent to one Red short of 6 Blue
*
Adding one Red on the Left makes 6 Blue which is equal to 9 Red*


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## bellenuit (13 November 2016)

pixel said:


> *We have to add one Red ball on the left and 8 Red balls (to make it 9) on the right.*
> 
> Calculation is simple:
> 4R + 2B = 2Y  =>  6Y = 6B + 12R
> ...




If you add one red ball on the left and 8 on the right, why bother adding the one on the left at all? Why not just add 7 red balls on the right? The one extra red on each side is superfluous as they are the same weight.


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## pixel (13 November 2016)

bellenuit said:


> If you add one red ball on the left and 8 on the right, why bother adding the one on the left at all? Why not just add 7 red balls on the right? The one extra red on each side is superfluous as they are the same weight.




LOL
you can be quite infuriating when you're right.
Especially when it didn't occur to me immediately :


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## bellenuit (13 November 2016)

pixel said:


> LOL
> you can be quite infuriating when you're right.
> Especially when it didn't occur to me immediately :




In any case your are correct. The "net" is you add 7 Red balls to the right, which brings the number to 8 Red balls on that side. 

However, I just noticed an error in your calcs.

Where you say subtraction gives:
1Y = 3B - 1R and 4B = 6R or 1B = 2/3 R

should read:
1Y = 3B - 1R and 4B = 6R or *1B = 1.5 R*

Then 3 B + 1 Y = 4.5 R + 4.5 R - 1R = 8 R

So 8 Red in total, meaning you add 7 Red.

I assume you just got it backwards (you meant to type 1R = 2/3B), so congrats anyway.


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## pixel (14 November 2016)

bellenuit said:


> In any case your are correct. The "net" is you add 7 Red balls to the right, which brings the number to 8 Red balls on that side.
> 
> However, I just noticed an error in your calcs.
> 
> ...




yes, that was indeed a typo. 
It was actually meant to read 1R = 2/3 B, 
and as I had 5 1/3 Blue on the Left, I added a Red to get to the integer number of 6 Blues or 9 Red.


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## Craton (27 September 2020)

What, no answers to the Cat Puzzle?


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## cynic (27 September 2020)

7 days. Choose either door 6 during the first 2 days and then decrement or choose door 2 during first 2 days and then increment.

Edit: I think I have gotten this wrong. Will try again shortly.


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## Craton (28 September 2020)

Yep, a tricky one. IMHO catching the cat starts with opening the door at one end for two consecutive days before moving onto the next adjacent door.

If the cat is behind door 2 and you open door 1 twice, then yes I think it is 7 days but, this is one smart moggy...


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## cynic (28 September 2020)

Craton said:


> Yep, a tricky one. IMHO catching the cat starts with opening the door at one end for two consecutive days before moving onto the next adjacent door.
> 
> If the cat is behind door 2 and you open door 1 twice, then yes I think it is 7 days but, this is one smart moggy...



If you start one door in from the end and do that twice then you know that the moggy cannot have been in the extreme end.

The problem is then what to do next because the moggy could be about to move into the door just checked twice from the opposite direction.

Anyway the moggy is alternating between odd and even numbered doors so that provides another angle that may merit consideration.
I.e. if behind odd numbered door on day one then will be behind odd numbered door on day three,five etcetera and even numbered door on the intervening days.


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## bellenuit (29 September 2020)

If you don't want a hint, don't read the following:

_Try with a corridor of just 3 doors. The answer to that is trivial. Then try with a corridor of just 4 doors. Solve that and then go to 5 doors. You might see a pattern emerge._


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