soo are there ANY smart people out there??

Question

you have 22 coins which all weigh the same and 2 coins which are heavier than the rest, but equal in weight to each other. You have a balance which can compare two groups of coins. What is the smallest number of weighings necessary to find the two heavy coins?

Answer 1

ok, I did some more thinking and maybe the answer is as low as 6.

24 coins, two are heavy. Break into three groups of 8. Weigh group A vs B and B Vs C.

Two weighs done and we will have either two stacks of 8 each with one heavy coin or one stack of 8 with two heavy coins.

With one coin in eight, weigh 1,2,3 vs 4,5,6 and then 1 vs 2, for two more weighs per stack of 8, or a total of six weighs to identify both coins.

Of if two heavy in 8 coins, then weigh 4 Vs 4 and then 1,2 vs 3,4 in both stacks of 4, again for a total of six weighs.

Answer 2

Hi. This is tricky. Break the coins into two groups of 12 and weigh one against the other. Since you know the two "odd" coins are heavier they will either balance or eliminate half the coins and identify which group has the heavier ones. If you are lucky (or we are lucky) we have one group with 12 equal weight coins to use as standards. Take the other group and break them into groups such that each group contains a unique number of coins. (i.e. 3, 4 & 5). Weigh each group against the standards. (3 v 3, 4 v 4, 5 v 5). Either 1 or 2 will be heavy. The heavy groups contain the heavy coins. OK, I'm spent.

Answer 3

Darn, This is a hard problem, do you have to weigh them?

the simpliest way to do this in my opinion is to put all the coins in water. the ones that sink first are the heavy ones. but let me try this.

I'll give it a stab, 11

Answer 4

what kind of question is that..... you can't have coins that weigh the same, but some are heavier.....

Answer 5

3, I read this some where...