**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 Continue reading “Problem of the Week 6 – Divisibility by 6”