# 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 3 – Largest Possible Value

Problem

The number $25ab$ where $a$ and $b$ are digits is divisible by $6$. What is the largest possible value of $a + b$?

Solution

A number is divisible by 6 if it is divisible by 2 and it is divisible by 3. For a number to be divisible by 2, it must be even. Therefore, $b$ must be even. So, the possible values of b are, 0, 2, 4, 6, and 8.

A number is divisible by 3 if the sum of the digits are divisible by 3. Therefore, we must be sure that in all the cases, the numbers are divisible by 3. Continue reading