AMC 10 · 2023 · #9

Grade 6 arithmetic
perfect-squaresdifference-of-squaressequences-arithmeticlinear-equations-one-var pattern-recognitioneasier-related-problemidentify-subproblems ↑ Prerequisites: perfect-squaresdifference-of-squares
📏 Short solution 💡 2 insights
Problem
Two consecutive positive perfect squares like 16 and 25 have difference 9. Count how many pairs of consecutive positive perfect squares have a difference that is at most 2023.

Pick an answer.

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

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Tool #9 (Easier Problem) says: don't fight (n+1)² - n² in the abstract — compute a handful of small cases and look. Tool #5 (Pattern) catches the result immediately: the differences are 3, 5, 7, 9, 11, … — the odd numbers in order. Tool #1 (Draw a Diagram) backs this up visually: lay n² as an n × n array of dots and growth to (n+1)² adds an L-shaped border of 2n + 1 dots. So the n-th difference is exactly 2n + 1, and the question turns into the one-line inequality 2n + 1 ≤ 2023, i.e. n ≤ 1011. Count the integers from 1 to 1011. No formal algebra needed — the pattern + the L-shape picture do the lifting.

1STEP 1

Compute the gap between the two squares for the first few n and look for a pattern.

n & 1 & 2 & 3 & 4 & 5 & 6 ; (n+1)² - n² & 3 & 5 & 7 & 9 & 11 & 13
2STEP 2

The gaps 3, 5, 7, 9, … are the odd numbers in order, so the n-th gap is 2n + 1.

(n+1)² - n² = 2n + 1
3STEP 3

The condition becomes 2n + 1 ≤ 2023, so subtract 1 and halve to get n ≤ 1011.

2n + 1 ≤ 2023 → 2n ≤ 2022 → n ≤ 1011
4STEP 4

Each integer n from 1 to 1011 gives one valid pair, so the number of pairs is 1011.

Number of pairs = 1011 - 1 + 1 = 1011 → (B)
Answer
1011
Three checks. (1) Endpoint: when n = 1011, the difference is 2(1011) + 1 = 2023, which is allowed since the problem says ≤ 2023, so n = 1011 counts — confirming the boundary. (2) Endpoint: when n = 1012, the difference is 2025 ＞ 2023, so n = 1012 correctly fails. (3) Magnitude: the differences are roughly 2n, so the cutoff n should be roughly 2023 / 2 ≈ 1011 — matches. Eliminate distractors: (A) 674 is just 2023 / 3, the wrong divisor; (C) 1010 is the off-by-one error that drops the n = 1011 endpoint; (D) 2019 and (E) 2017 are red herrings using 2023 itself.
💡Key takeaway

This AMC 10 problem only needs Grade 6 inequalities you already know — the gap between consecutive squares is always the next odd number 2n + 1, so the question becomes 2n + 1 ≤ 2023, giving n ≤ 1011, hence 1011 pairs.