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2023 AMC 10B Problem 9

2023 AMC 10B Problem 9AMC 10 Step by Step

Name: Date: September 13, 2026

Number Properties · difficulty 2 of 5 · about 2 min · position P1-10

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

A)
$674$
B)
$1011$
C)
$1010$
D)
$2019$
E)
$2017$

Answer

Key insight

Consecutive squares differ by (n+1)^2 - n^2 = 2n+1, so the condition 2n+1 <= 2023 means n = 1, 2, ..., 1011.

Solution

A pair of consecutive positive squares is $n^2$ and $(n+1)^2$ with $n \ge 1$ . Their difference is
$(n+1)^2 - n^2 = 2n + 1.$
We need $2n + 1 \le 2023$ , i.e. $n \le 1011$ . Each $n = 1, 2, \dots, 1011$ gives exactly one pair (and $n = 1011$ works since $2 \cdot 1011 + 1 = 2023$ ), so there are $1011$ pairs.

The answer is $\boxed{\textbf{(B)}\ 1011}$ .

Why this works

The difference of consecutive squares is the odd number $2n+1$ , so "difference at most $2023$ " is a linear condition on $n$ and the count is immediate. Always name the pair by a single index ( $n$ ) so that counting pairs becomes counting integers in a range.

The trap

Using a strict inequality or starting from n = 0, which shifts the count to 1010 or 1012.

Common mistakes

Using a strict inequality or starting from n = 0, which shifts the count to 1010 or 1012.
Counting odd numbers up to $2023$ ( $1012$ of them, including $1$ ) without noticing that difference $1$ would require $0^2$ , which is not positive.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta · Bound the quantity above/below or estimate to pin it down