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2023 AMC 10B Problem 9

Problem 9 of 25EasierAlgebraNumber TheoryCounting & Probability

The numbers 1616 and 2525 are a pair of consecutive positive perfect squares whose difference is 9.9. How many pairs of consecutive positive perfect squares have a difference of less than or equal to 2023?2023?

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Solution

Consecutive squares k2k^2 and (k+1)2(k+1)^2 differ by (k+1)2−k2=2k+1.(k+1)^2 - k^2 = 2k + 1. We need 2k+1≤2023,2k + 1 \le 2023, which gives k≤1011.k \le 1011. So kk runs 1,2,…,1011,1, 2, \ldots, 1011, for 10111011 pairs. Thus, B is the correct answer.

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Concepts: perfect square · difference of squares · counting integers in a range

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.