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Weekly Problem 22 - 2007

Stage: 2 and 3 Challenge Level: Challenge Level:1
7

The four youngest must receive at least 1 + 2 + 3 + 4 = 10 sweets, so the oldest receives at most 10 sweets. Also, as 5 + 4 + 3 + 2 + 1 = 15, which is less than twenty, the oldest must receive at least 6 sweets. Systematically listing the different ways of sharing the sweets gives (10, 4, 3, 2, 1), (9, 5, 3, 2, 1), (8, 6, 3, 2, 1), (8, 5, 4, 2, 1), (7, 6, 4, 2, 1), (7, 5, 4, 3, 1) and (6, 5, 4, 3, 2).

This problem is taken from the UKMT Mathematical Challenges.

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