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”

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 “Problem of the Week 3 – Largest Possible Value”