2023 AMC 10B Problems/Problem 9

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

$\text{(A)}\ 674 \qquad \text{(B)}\ 1011 \qquad \text{(C)}\ 1010 \qquad \text{(D)}\ 2019 \qquad \text{(E)}\ 2017$

Solution

Let m be the sqaure root of the smaller of the two perfect squares. Then, $(m-1)^2 - m^2 = (2m+1)(1) = 2m+1 \le 2023$ . Thus, $m \le 1011$ . So there are 1011 numbers that satisfy the equation. $\boxed{\text{B}}$

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2023 AMC 10B Problems/Problem 9

Solution