2023 AMC 10B Problem 9

Below is the professionally curated solution for Problem 9 of the 2023 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 10B solutions, or check the answer key.

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

Difficulty rating: 1310

The numbers $16$ and $25$ are a pair of consecutive positive perfect squares whose difference is $9.$ How many pairs of consecutive positive perfect squares have a difference of less than or equal to $2023?$

$674$

$1011$

$1010$

$2019$

$2017$

Solution:

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

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