2006/07 AHSMC Part 1 Problem 14

Problem 14 of 16

Colleen used a calculator to compute $\dfrac{a + b}{c}$ , where $a$ , $b$ and $c$ are positive integers. She pressed $a$ , $+$ , $b$ , $/$ , $c$ and $=$ in that order, and got the answer $11$ . When she pressed $b$ , $+$ , $a$ , $/$ , $c$ and $=$ in that order, she was surprised to get a different answer $14$ . Then she realized that the calculator performed the division before the addition. So she pressed $($ , $a$ , $+$ , $b$ , $)$ , $/$ , $c$ and $=$ in that order. She finally got the correct answer, which is

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Solution

We have $a + \dfrac{b}{c} = 11$ and $b + \dfrac{a}{c} = 14$ . Adding the two equations yields $(a + b)\dfrac{c + 1}{c} = 25$ or $(a + b)(c + 1) = 25c$ . Since $c + 1$ and $c$ are relatively prime, $c + 1$ must divide $25$ . Hence $c = 4$ or $c = 24$ . If $c = 24$ , then $a + b = 25$ but at least one of $\dfrac{a}{c}$ and $\dfrac{b}{c}$ is not an integer. Hence $c = 4$ , $a + b = 20$ and $\dfrac{a + b}{c} = 5$ . The answer is (b).

University of Alberta

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