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2021/22 AHSMC Part 1 Problem 13

Problem 13 of 16

Let mm, nn be positive integers such that 1m+1n=1994\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{19}{94}. The value of m+nm + n is equal to

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Solution

We may assume that m≥nm \ge n, hence 19n94=nm+1∈(1,2]  ⟺  n∈(9419, 18819]  ⟺  n∈{5,6,7,8,9}. \frac{19n}{94} = \frac{n}{m} + 1 \in (1, 2] \iff n \in \left(\frac{94}{19},\ \frac{188}{19}\right] \iff n \in \{5,6,7,8,9\}. Only n=5n = 5 is possible, which gives m=470m = 470, and hence n+m=475n + m = 475. The answer is (D).

University of Alberta