You have 22 coins which all weigh the same and 2 coins that are heavier than the rest, but equal in weight to each other. You have a balance that can compare two groups of coins. What is the smallest number of weighings necessary to find the two heavy coins?
Sep 24th 2010, 02:28 PM
This is what I have so far....
We know we have 24 coins in all, but should I make two or three piles to weigh? Should I take two piles of 12 or three piles of 8? And where do I go from here?
Any insight would be nice....thank you in advance!