2023 AMC 10B Problems/Problem 9

Revision as of 17:39, 15 November 2023 by E is 2.71828 (talk | contribs) (Solution 1)

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 1

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 $\boxed{\text{(B)}1011}$ numbers that satisfy the equation.

~andliu766