AMC 10 · 2023 · #9
Grade 6 arithmeticProblem
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?
Pick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $2023$.
Givens: A pair means $n^2$ and $(n+1)^2$ for some positive integer $n \ge 1$; Their difference is $(n+1)^2 - n^2$; Sample: $4^2 = 16$ and $5^2 = 25$ have difference $9$; We want all pairs whose difference is $\le 2023$; Answer choices: (A) $674$, (B) $1011$, (C) $1010$, (D) $2019$, (E) $2017$
Unknowns: The number of values of $n$ (so the number of pairs) for which the difference is at most $2023$
Understand
Restated: 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 $2023$.
Givens: A pair means $n^2$ and $(n+1)^2$ for some positive integer $n \ge 1$; Their difference is $(n+1)^2 - n^2$; Sample: $4^2 = 16$ and $5^2 = 25$ have difference $9$; We want all pairs whose difference is $\le 2023$; Answer choices: (A) $674$, (B) $1011$, (C) $1010$, (D) $2019$, (E) $2017$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #9 Solve an Easier Related Problem, #1 Draw a Diagram
Tool #9 (Easier Problem) says: don't fight $(n+1)^2 - n^2$ in the abstract — compute a handful of small cases and look. Tool #5 (Pattern) catches the result immediately: the differences are $3, 5, 7, 9, 11, \ldots$ — the odd numbers in order. Tool #1 (Draw a Diagram) backs this up visually: lay $n^2$ as an $n \times n$ 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 $2n + 1 \le 2023$, i.e. $n \le 1011$. Count the integers from $1$ to $1011$. No formal algebra needed — the pattern + the L-shape picture do the lifting.
Execute — Answer: B
3.OA.D.9 Step 1 - Compute the differences for the first few values of $n$ to look for a pattern.
- $n = 1$: $2^2 - 1^2 = 3$.
- $n = 2$: $3^2 - 2^2 = 5$.
- $n = 3$: $4^2 - 3^2 = 7$.
- $n = 4$: $5^2 - 4^2 = 9$.
- $n = 5$: $6^2 - 5^2 = 11$.
💡 Shrinking to small cases reveals the pattern — Grade 3 arithmetic pattern finding.
4.OA.C.5 Step 2 - The differences $3, 5, 7, 9, 11, 13, \ldots$ are the odd numbers starting at $3$, so the $n$-th difference is $2n + 1$.
- Picture-check: growing an $n \times n$ square into an $(n+1) \times (n+1)$ square adds a backwards-L border made of $n$ dots on the right, $n$ dots on the top, and $1$ corner dot — that is $n + n + 1 = 2n + 1$ new dots.
💡 Reading off the rule for the $n$-th difference — Grade 4 generating a number pattern by a rule.
6.EE.B.8 Step 3 - The condition the problem asks for becomes $2n + 1 \le 2023$.
- Subtract $1$ from both sides, then halve.
💡 Solving a one-step inequality of the form $x \le c$ — Grade 6 inequality reasoning.
3.OA.A.3 Step 4 - $n$ must be a positive integer with $n \le 1011$.
- Each such $n$ gives one valid pair $(n^2, (n+1)^2)$, so count the integers from $1$ to $1011$ inclusive.
💡 Counting consecutive integers from $1$ to $1011$ — Grade 3 word-problem counting.
3.OA.D.9 Compute the differences for the first few values of $n$ to look for a pattern. $ 4.OA.C.5 The differences $3, 5, 7, 9, 11, 13, \ldots$ are the odd numbers starting at $3$ 6.EE.B.8 The condition the problem asks for becomes $2n + 1 \le 2023$. Subtract $1$ from 3.OA.A.3 $n$ must be a positive integer with $n \le 1011$. Each such $n$ gives one valid Review
Reasonableness: Three checks. (1) Endpoint: when $n = 1011$, the difference is $2(1011) + 1 = 2023$, which is allowed since the problem says $\le 2023$, 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 $2023 / 2 \approx 1011$ — 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.
Alternative: Tool #13 (Convert to Algebra): set $D = (n+1)^2 - n^2$ and expand $(n+1)^2 = n^2 + 2n + 1$, giving $D = 2n + 1$. Then solve $2n + 1 \le 2023$ to get $n \le 1011$. Same logic, but the algebraic expansion is heavier than the L-shaped-border picture for the same one-line formula.
CCSS standards used (min grade 6)
3.OA.D.9Identify arithmetic patterns and explain using properties of operations (Computing the first few differences $3, 5, 7, 9, 11, \ldots$ and noticing they form the odd numbers.)4.OA.C.5Generate a number or shape pattern following a given rule (Stating the rule that the $n$-th difference is $2n + 1$, verified by the L-shaped border picture.)6.EE.B.8Write an inequality of the form $x > c$ or $x < c$ and graph on a number line (Translating the condition into $2n + 1 \le 2023$ and solving for $n \le 1011$.)3.OA.A.3Solve multiplication and division word problems within 100 (Counting that the integers from $1$ to $1011$ inclusive number $1011$ — one count per valid pair.)
⭐ 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 $2n + 1 \le 2023$, giving $n \le 1011$, hence $1011$ pairs.
⭐ 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 $2n + 1 \le 2023$, giving $n \le 1011$, hence $1011$ pairs.