2023 AMC 10B Problem 9

Question

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

Consecutive squares

k^2

and

(k+1)^2

differ by

2k+1

Solve

2k + 1 \le 2023

for positive integers

k

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.

Problem 9 in other years