# Maths prize for 18 December: 12 billiard balls

You have 12 billiard (or snooker) balls, all the same except one which is slightly heavier – or it may be slightly lighter – than the rest.

You have scales which you can use to compare balls (or groups of balls) with each other.

How can you identify the odd ball by using the scales only three times?

Thanks to Hamse Adam for bringing us this puzzle. It doesn’t call for fancy maths, but it does call for a bit of thinking, so we gave online hints to help you. The first hint is at:

https://mathsmartinthomas.wordpress.com/2015/11/29/the-12-billiard-balls-puzzle-hint-1/

Wei-Kong Mao, Mai Nguyen, and Zion Mills won prizes.

This is Wei-Kong’s solution:

Number the balls 1 to 12.

Let’s split the balls into three groups: A,B and C

Group A Group B. Group C
[1 2 3 4]. [5 6 7 8] [9 10 11 12]

Let’s decide to weigh group A and B
There are three possible outcomes from using the scales once. Either A and B balance or A > B or A < B.

If A and B balance:

Act 2a: now we use the scale to balance any three balls from group A or B -because they are all the same weight – with three balls from group C e.g
1,2,3 vs 9, 10, 11

Act 3a: if 1,2,3 vs 9, 10, 11 balances the means the odd ball is ball 12 so then we use the last usage of the scale is check whether it lighter or heavier.

Act 2b: But if 9, 10, 11 is heavier that means the odd ball is is heavy and is either 9 10 or 11

Act 3b: then we use the scale a final time with 9 vs 10 ; if they balance , 11 is the odd heavy. If they don’t balance the heavier one is the odd one.

Act 2c: If 9,10,11 is light then they contain a light ball

Act 3c: weigh ball 9 vs 10 if they balance ball 12 is odd light ball and if they don’t balance the lighter ball is the odd light ball.

Suppose they don’t balance, for example A > B. This means either A contains a heavy ball or B contains a light one.

Act 2a: let’s take a ball from group A and two balls from group B, and weigh those three against one ball each from A, B and C
e.g 1,5,6 vs 2,7,9

Act 3a: If they balance that means 8 is an odd light ball or 3 or 4 is a heavy ball. Then weigh ball 3 and 4. If they balance then 8 is an odd light ball; if they don’t balance, then the heavier one is the heavy odd ball

Act 2b: What if they don’t balance? Suppose 2,7,9 is heavy. This means 2 is the heavy odd ball or 5 or 6 is the light odd ball. (We know that 9 is not an odd ball, because it’s in group C. We can do a similar process to when either 8 is light or 3 or 4 is heavy.

Act 2c: If 2,7,9 is lighter that means ball 1 is the heavy odd ball or ball 7 is the light odd ball. Then weigh ball against any other ball. If they balance, then ball 1 is the odd heavy ball. Else 7 is the odd light ball.

You would do the same for A < B.

*Weigh 2*: compare 1-2-5 and 3-4-6

In every case, we have narrowed it down to three or two possibilities, and we can use the sort of method Wei-Kong describes to find the actual case among those two or three with the third weighing.