The Chicago area business services firm asked this of an applicant for an operations associate position in March 2013.
This question is a more difficult variation on the classic "balance problem." The procedure depends on whether you have a normal coin for reference.
One user from Math forum.org offers his solution:
"So all we have to do is to distribute the 12 coins over the scales of
the three measurements in such a way that no coin participates in the
three measurements in the same way (or mirrored) as any other coin.
The distribution below is one of many possible distributions that
fulfills this requirement:
1, 2, 7, 10 against 3, 4, 6, 9
1, 3, 8, 11 against 2, 5, 6, 7
2, 3, 9, 12 against 1, 4, 5, 8
If the measurements result unequally, then it can be seen from this
distribution that coin 8 is lighter than the other coins. No other
coin can explain the unequal outcome."
Number people 1, 2 , 3 ,4 , 5 meaning their ordinal age; there are five ways they can sit around a round table in increasing order of age: [ 1 2 3 4 5 ] [2 3 4 5 1] etc. There are 5 x 4 x 3 x 2 x 1 possible seating arrangements around the table, so dividing 5 by 5 x 4 x 3 x 2 we get probability = 1/24