# Coin Collection

When coins are put into piles of six 3 remain and in piles of eight 7 remain. How many remain when they are put into piles of 24?

Tom has a collection of more than $24$ coins. When he puts the coins in piles of $6$, there are $3$ coins remaining. When he puts the coins in piles of $8$, there are $7$ coins remaining. How many coins remain when he puts the coins in piles of $24$?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

**Answer**: 15

**Counting backwards**

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Multiples of both 6 and 8 are multiples of 24

**Using number facts**

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.