If B and C are digits, and 9BC is a 3-digit number that is divisible

Solution:

Recall that a number is divisible by 3 if the sum of its digits is divisible by 3. Since 9 is divisible by 3, B + C must be divisible by 3. For example, if B = 6 and C = 0, then B + C = 6 is divisible by 3 and we see that 0 could be a possible product of B and C. However, let’s prove that all the other choices can’t be a possible product of B and C.

B) Since 4 = 2 x 2 = 1 x 4, we see that B + C is either 2 + 2 = 4 or 1 + 4 = 5, neither of which is divisible by 3.

C) Since 15 = 3 x 5, we see that B + C is 3 + 5 = 8, which is not divisible by 3.

D) Since 16 = 4 x 4 = 2 x 8, we see that B + C is either 4 + 4 = 8 or 2 + 8 = 10, neither of which is divisible by 3.

E) Since 21 = 3 x 7, we see that B + C is 3 + 7 = 10, which is not divisible by 3.

Answer: A

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