2020/21 AHSMC Part 1 Problem 4

Problem 4 of 16

$1a5$ and $6b9$ are two $3$ -digit numbers, where $a$ and $b$ are the middle digits of the two numbers. If $6b9 - 1a5 = 454$ and $6b9$ is divisible by $9$ , then $a + b$ is equal to

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Solution

The rule for divisibility by $9$ says that a number is divisible by $9$ if the sum of its digits is divisible by $9$ . Hence, $6 + b + 9 = 15 + b$ is divisible by $9$ . Therefore $b = 3$ and hence $6b9 = 639$ .

The equality $639 - 1a5 = 454$ gives $a = 8$ . Therefore $a = 8$ , $b = 3$ and $a + b = 11$ . The answer is (E).

University of Alberta

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