Sunday Times Teaser 2500 – No Title

by Peter Harrison

A well-known puzzle asks: “If among 12 snooker balls one is a different weight, how can the rogue ball be identified — together with deciding whether it is heavier or lighter — in three weighings on a balance?”

Recently I faced a very similar problem of finding a rogue ball among a consignment of 39 identical- looking balls — and deciding if it was heavier or lighter. I had at my disposal a two-pan balance.

How many weighings did I need?

One Comment Leave one →
1. Puzzles of this kind (with coins rather than snooker balls) are  fairly well known with quite extensive coverage on the internet.  In situations where there is exactly one bad coin but it is not known whether it is lighter or heavier than normal, it is possible to identify it using $$N$$ weighings within at most $$(3^N-3)/ 2$$ coins.  A methodical way of doing this has been developed by Jack Wert and is described on the Cut the Knot web site. This is the approach that will be described here for the 39 coin case. The first step is to divide the 39 coins into three groups of 13 and then to further arrange each such group into three piles of 9, 3 and 1 coin.   Two of these groups of 13 coins are placed on the balance and the third is placed on the table as shown above.   Note that the scales in this first weighing might or might not be in balance – we are looking for changes in the position of the scales when we move coins.

In the second weighing we exchange one of the 9 coin piles on the scales with the 9 coin pile on the table.  If the balance doesn’t change then we know that all three 9 coin piles weigh the same so we can ignore them and consider the remaining 12 coins.

We now repeat the process by exchanging one of the 3 coin piles on the scales with the one on the table, noting in this third weighing whether the balance has changed.   If it hasn’t, we know that all three coin piles weigh the same so we can ignore them and repeat the process again with the 3  remaining piles of single coins.

By exchanging a one coin pile on the scales with the one on the table, a fourth weighing  will identify and ‘weigh’ the bad coin.

If the balance does change and it was balanced but now isn’t, the bad coin is in the 3 coin pile we moved onto the scales and we also now know whether it is lighter or heavier; if it wasn’t balanced but now is, the bad coin is in the three coin pile we moved onto the table and we again know from the pan it was moved from whether it is lighter or heavier.

We can now work with these three coins by weighing 2 against each other in a fourth weighing.   If the scales balance it is the coin that we didn’t weigh that is bad and we already know whether it is lighter or heavier.  It they don’t balance we know which pan the holds the bad coin since we know whether it is lighter or heavier.

If the balance does change and it is now balanced we know that the 9 coin pile we moved onto the table contains the bad coin and we know whether it is lighter or heavier since we know the pan from which we moved it. If on the other hand it was balanced but now isn’t, the 9 coin pile we moved onto the scales contains the bad coin and we know from the position of the scales whether it is lighter or heavier.  So we can now clear the scales and work with these nine coins as follows.

We divide them into 3 groups of 3 coins and weigh two of the groups against each other in a third weighing. If the scales balance it is the group of three on the table that contains the bad coin. If the scales don’t balance we know which side contains the bad coin because we already know whether it is lighter or heavier.

We now work with these three coins by weighing 2 against each other in a fourth weighing as described earlier.