# Problem of the Week 6 – Divisibility by 6

**Problem**

The number 25*a* where *a* is a digit is divisible by 6. What is the largest possible value of *a*?

*Solution*

A number is divisible by 6 if it is both divisible by 2 and divisible by 3.

First, a number if divisible by 2 if it is even. Now, for 25*a* to be even, *a* must be even. Since *a* is even, it could be one of the following numbers:* *0, 2, 4, 6 or 8.

Second, a number is divisible by 3 if the sum of its digit is divisible by 3. So, we substitute the possible numbers above, add the digits, and see if they are divisible by 3.

*For a* = 0

2 + 5 + 0 = 7

Is 9 divisible by 3? NO.

*For a* = 2

2 + 5 + 2 = 9

Is 9 divisible by 3? YES.

*For a* = 4

2 + 5 + 4 = 11

Is 11 divisible by 3? NO.

*For a* = 6

2 + 5 + 6 = 13

Is 13 divisible by 3? NO.

*For a* = 8

2 + 5 + 8 = 15

Is 15 divisible by 3? YES.

Therefore, for divisibility by 3, only two digits qualify: *a* = 2 and *a* = 8. And since we are looking for the largest possible value of *a*, the correct answer is ** a = 8**.

So, the number is 258.

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