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 

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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