ASF Spoilers Thread

[cynic]

(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.

 

[bellenuit]

(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.

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.

 

[pixel]

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

 

[bellenuit]

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

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.

 

[pixel]

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 :

 

[bellenuit]

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.

 

[pixel]

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.

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.

 

[Craton]

What, no answers to the Cat Puzzle?

 

[cynic]

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.

 

[Craton]

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...

 

[cynic]

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.

 

[bellenuit]

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.