6 ms·
How do you solve this? ABC, one of them is different weight than the other two (either more or less) Possible ways to weigh them: A v B A not enough info
by emilfihlman 8y ago
How do you solve this?
ABC, one of them is different weight than the other two (either more or less)
Possible ways to weigh them:
A v B
A not enough information
E C is odd one out
B not enough information
A v C
A not enough information
E B is odd one out
C not enough information
B v C
B not enough information
E A is odd one out
C not enough information
AB v C
AB not enough information
E C is odd one out
C C is odd one out
AC v B
AC not enough information
E B is odd one out
B B is odd one out
BC v A
BC not enough information
E A is odd one out
A A is odd one out
In all possible ways to weight 3 coins, all have uncertainty which means you cannot deduce which is the odd one out.
- scrooched_moose 8y agoYour OP stated one is lighter, not one is different. It's easier than the 12-coin one different puzzle, but introduces the ternary concept well enough to get you started on the harder problem (there are some other complications to it as well). Weigh 2 coins. If the scales move you've found the lighter coin. If the scales balance, it's the third coin you didn't weigh.
- emilfihlman 8y agoAh I got the original (12 coins, one is lighter or heavier) mixed up with 3 coins (one is lighter).
- deleted 8y ago[deleted]
- grive 8y agoLabel each coin, divide in three stacks each time, and shuffle meaningfully each time the subsets, to remove uncertainty.