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Weekly Problem 15 - 2014

Stage: 3 and 4 Challenge Level: Challenge Level:1

A number is a multiple of 6 precisely when it is both a multiple of 2 and of 3. To be a multiple of 2, it will need to end with an even digit; i.e. 0 or 2. To be a multiple of 3, the sum of the digits has to be a multiple of 3.

If it ends with 0, the sum of the other two digits must be a multiple of 3; and only $3 = 1 + 2$ or $6 = 1 + 5$ are possible. That gives the numbers 120, 210, 150, 510.
If it ends with 2, the sum of the others must be $1 = 0 + 1$ or $4 = 1 + 3$. That gives 102, 132 and 312.

Hence 7 of these numbers are divisible by 6.

This problem is taken from the UKMT Mathematical Challenges.

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