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2013/14 AHSMC Part 1 Problem 7

Problem 7 of 16

Let aa and bb be positive integers such that ab<100ab < 100 and ab>2\dfrac{a}{b} > 2. Denote the minimum possible value of ab\dfrac{a}{b} by mm. Then we have

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Solution

We have 100>ab>b(2b+1)100 > ab > b(2b+1) so that b≤6b \le 6. Thus the minimum value of mm is 2+16=2.1666…2 + \dfrac{1}{6} = 2.1666\ldots. The answer is (B).

University of Alberta