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ASF Puzzles & Conundrums Thread

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You have a box containing a number of red and blue balls. You also have an unlimited supply of red and blue balls outside the box at your disposal.

You take two balls from the box at random. If they are the same colour you put back in a red ball. If they are different colours you put back in a blue ball. Thus after each go, you reduce the number of balls in the box by 1 and will eventually have just 1 ball in the box which ends the game.

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.
 
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.

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.
 
Just to keep it moving along.

There are 4 closed boxes each containing a ball of a different colour (one ball per box). There is a competition to see who can guess the correct coloured ball in each box. 123 people participate.

When the boxes are opened, it turns out that:
43 people have guessed none of the box contents correctly,
39 have guessed 1 box correctly,
31 have guessed 2 boxes correctly.

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

(Note: it should be obvious from the figures, but guessing 2 correctly means exactly 2. Ditto for 3 and 4. So no arguments such as if you guessed 2 correctly you also must have guessed 1 correctly. They are not cumulative.
 

43+39+31=113
Leaves 10 to get them all right.
 
pixel said:
43+39+31=113
Leaves 10 to get them all right.
Click to expand...

Pixel, the spoiler thread is the place to post your answers, it was set up so people dont see the answers when they open this thread - before they have had a chance to solve them.

(makes me think about a potentially more elegant solution, some forums have a 'spoiler text' option that means you can only read that segment of the post when you hover the mouse over it - that would save having a separate thread but also hide the answers people posted.)
 
A hard one to put your brain in top gear. It does require some math though, but not much more than knowing the circumference of a circle is 2πR (π being the mathematical constant pi, in case it doesn't display correctly on your device).

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.
 
I'm updating the last paragraph of this question as it was phrased badly and may lead people astray.

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 is very smart and always tries to get to that point on the shore that is closest to where you currently are? If so, explain how.
 


Last Slylock I got.

If you enjoyed, I highly recommend buying the books. For kids/partner/yourself(?)
 
Denise's Birthday Puzzle (Cheryl Part 2)

Albert, Bernard and Cheryl became friends with Denise, and they wanted to know when her birthday is. Denise gave them a list of 20 possible dates.

17 Feb 2001, 16 Mar 2002, 13 Jan 2003, 19 Jan 2004

13 Mar 2001, 15 Apr 2002, 16 Feb 2003, 18 Feb 2004

13 Apr 2001, 14 May 2002, 14 Mar 2003, 19 May 2004

15 May 2001, 12 Jun 2002, 11 Apr 2003, 14 Jul 2004

17 Jun 2001, 16 Aug 2002, 16 Jul 2003, 18 Aug 2004

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.
Bernard: I still don’t know when Denise’s birthday is, but I know that Cheryl still does not know.
Cheryl: I still don’t know when Denise’s birthday is, but I know that Albert still does not know.
Albert: Now I know when Denise’s birthday is.
Bernard: Now I know too.
Cheryl: Me too.

So, when is Denise’s birthday?

To clarify: neither Albert, Bernard or Cheryl know anything else at the start apart from the fact that Albert has been told the month, Bernard the day (meaning the number of the day), and Cheryl the year.
 
An Saudi sheikh tells his two sons to race their camels to a distant city to see who will inherit his fortune. The one whose camel is slower wins. After wandering aimlessly for days, the brothers ask a wise man for guidance. After hearing his advice, they jump on the camels and race to the city as fast as they can. So what advice did the wise man offer to them?
 
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?
 
Not as easy as it looks.......

Interesting Numbers

Give me a 7 digit number with the following properties:

The 1st (leftmost) digit gives the number of digit 0s in the number.
The 2nd digit gives the number of digit 1s in the number.
The 3rd digit gives the number of digit 2s in the number.
The 4th digit gives the number of digit 3s in the number
etc. until finally ...
The 7th digit gives the number of digit 6s in the number.

For example, 1210 is a 4 digit number with those same properties

If you are adventurous, you can also gives answers for 8, 9 and 10 digit numbers.

It should be obvious that there isn't an answer for every possible number of digits. For example, there isn't a 1 digit number that could have those properties.

And for those of you who have nothing better to do, give a 16 digit number (in hexadecimal) that has those properties. Hexadecimal, for those not familiar with it, is to the base 16 rather than base 10 of our decimal system. It uses A, B, C, D, E and F to represent 10 to 15 in the number scale.
 
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?
 

50?
 

Does not compute ?
 

To clarify, in case there is confusion. The initial marble has a 50% chance of being Red and a 50% chance of being Blue. This is in case someone is thinking that the chance of it being red or blue is 50% (25% each) and the chance of it being another colour is 50%.
 
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?

(Crucial) hint: You have been given enough information to solve this problem.

For those whose schooldays are a distant memory, the volume of a sphere is (4πr**3)/3 where π is Pi and r**3 means the radius r cubed.

 
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