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2008/09 AHSMC Part 1 Problem 9

Problem 9 of 16

The positive integers aa and bb are such that there are exactly 1010 integers greater than aa and less than bb, and exactly 10001000 integers greater than a2a^2 and less than b2b^2. The value of bb is

Answer choices

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Solution

We have b−a=11b - a = 11 and b2−a2=1001b^2 - a^2 = 1001, so that b+a=91b + a = 91. Hence 2b=1022b = 102 and b=51b = 51. The answer is (B).

University of Alberta