Challenge Level

Multiples of both 6 and 8 are multiples of 24

The number of coins in Tom's collection is 3 more than a multiple of 6 and also 7 more than a multiple of 8. The smallest number that satisfies both conditions is 15. The lowest common multiple of 6 and 8 is 24, so the conditions will also be met by numbers that exceed 15 by a multiple of 24, that is 39, 63, 87, etc. So when Tom puts his coins in piles of 24, 15 remain.

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.