Question

Let A and B be digits (that is, A and B are integers between 0 and 9 inclusive). If the product of the three-digit integers 2A5 and 13B is divisible by 36, determine the number of possible ordered pairs (A,B).

• A. 1
• B. 3
• C. 4

Answer 1

The correct option is C 4

For the product (2A5)(13B) to be divisible by 36, we need it to be divisible by both 4 and 9. Since 2A5 is odd, it does not contain a factor of 2.
Therefore, 13B must be divisible by 4.

For a positive integer to be divisible by 4, the number formed by its last two digits must be divisible by 4, i.e. 3B is divisible by 4, i.e. B = 2 or B = 6.

Case 1: B = 2

In this case, 132 is divisible by 3, but not by 9. Therefore, for the original product to be divisible by 9, we need 2A5 to be divisible by 3.

For a positive integer to be divisible by 3, the sum of its digits is divisible by 3, i.e. 2 + A + 5 = A + 7 is divisible by 3.

Therefore, A = 2 or 5 or 8.

Case 2: B = 6

In this case, 136 contains no factors of 3, so for the original product to be divisible by 9, we need 2A5 to be divisible by 9.

For a positive integer to be divisible by 9, the sum of its digits is divisible by 9, i.e. 2+A+5 is divisible by 9. Therefore, A is 2.

Therefore, the four possible ordered pairs are $\Rightarrow$(A, B) = (2,2), (8,2), (5,2), (2,6).