Problem of the Week 6 – Divisibility by 6

Problem

The number 25a 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 25a 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”