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Thank you for the many solutions to this problem. It was interesting to see that some of you presumed there had to be three cards in each pile which totalled 15. In fact, the question simply said there had to be three PILES, which makes the problem a little trickier than it looks at first.

Jessica and Ruby from Aldermaston C of E Primary School told us how they went about tackling the problem:

We tried to be systematic by looking for other ways to make a
number, for example 7. So we made 6+1, 3+4, 5+2 and 7 on its own
and put 8 with it each time. It was good because then we could get
more solutions to make 15. Here are our solutions to make 15
(reading across the table for the three piles):

2, 3, 9, 1 | 6, 5, 4 | 7, 8 |

3, 8, 4 | 6, 7, 2 | 9, 1, 5 |

5, 2, 8 | 1, 3, 4, 7 | 9, 6 |

6, 1, 8 | 5, 7, 3 | 2, 4, 9 |

Wilbury Primary School Mathletics Club also got the idea. Some of the solutions they found were the same as Jessica's and Ruby's, but here are their different solutions:

1, 2, 3, 4, 5 |
9, 6 | 7, 8 |

9, 1,5 | 7, 8 | 4, 3, 6, 2 |

9, 6 | 3, 5, 7 | 1, 8, 2, 4 |

So, in total Jessica, Ruby and the Mathletics Club at Wilbury have found seven different ways of putting the cards into three piles.

Then we asked you
if these were all the combinations possible. Alicia and William of
St Hugh's Woodhall Spa rose to the challenge, finding one more
combination:

9, 6 | 8, 4, 3 | 7, 5, 2, 1 |

and telling us why they thought they'd found them all:

We worked systematically swapping combinations of the same
value, ie 1 and 5 with 2 and 4. That way we took piles like (9, 1,
5), (8, 7), (6, 3, 2, 4) and got the new bunch of piles (9, 2, 4),
(8, 7), (6, 3, 1, 5). We did this for all the pairs, ones and
threes that made up the same value, and then got rid of the ones
we'd counted twice.

That makes eight ways altogether. Well done, Alicia and William! I think there might be one more to find ...