AMC 10 · 2024 · #15
Grade 8 number-theoryalgebraProblem
Let M be the greatest integer such that both M+1213 and M+3773 are perfect squares. What is the units digit of M?
Pick an answer.
AMC 10 2024 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: Find the largest integer $M$ such that $M + 1213$ and $M + 3773$ are both perfect squares. Then report the units digit of $M$.
Givens: $M + 1213 = a^2$ for some non-negative integer $a$; $M + 3773 = b^2$ for some non-negative integer $b$; $M$ is the greatest integer that satisfies both; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $6$, (E) $8$
Unknowns: The units digit of $M$
Understand
Restated: Find the largest integer $M$ such that $M + 1213$ and $M + 3773$ are both perfect squares. Then report the units digit of $M$.
Givens: $M + 1213 = a^2$ for some non-negative integer $a$; $M + 3773 = b^2$ for some non-negative integer $b$; $M$ is the greatest integer that satisfies both; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $6$, (E) $8$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems
The phrase "both ... are perfect squares" names two unknown squares with a fixed difference — exactly the trigger for Tool #13 (Convert to Algebra): give the two squares names $a^2$ and $b^2$, subtract to eliminate $M$, and the unknowns are now an integer pair $(a,b)$ with $b^2 - a^2 = 2560$. The difference of squares factors that into $(b-a)(b+a) = 2560$, turning the squares problem into a factor-pair problem. Tool #7 (Identify Subproblems) then splits the work cleanly: first find $a$ from the right factor pair (max $M$ comes from the most lopsided pair), then turn $a$ into the units digit using only the ones-digit arithmetic of $a^2 - 1213$.
Execute — Answer: E
6.EE.A.2 Step 1 - Name the two squares and subtract to remove $M$.
- Set $M + 1213 = a^2$ and $M + 3773 = b^2$.
- Subtracting the first from the second eliminates $M$ and leaves a fixed difference of squares.
💡 Writing both "perfect square" conditions as equations lets you cancel the unknown $M$ — the kind of move Grade 6 introduces when it teaches expressions with variables.
6.NS.B.4 Step 2 - Factor the difference of squares.
- The right side becomes a product of two integers $b-a$ and $b+a$.
- Note their sum is $2b$ and their difference is $2a$, so $b-a$ and $b+a$ must have the same parity.
- Since their product $2560$ is even, both factors must be even.
💡 Difference of squares is the algebra version of arranging $b^2 - a^2$ as an L-shaped strip whose two side lengths are $b-a$ and $b+a$. Same-parity is the integer-factorization constraint that comes for free.
4.OA.B.4 Step 3 - Subproblem A: pick the factor pair that maximizes $M$.
- From $a = \dfrac{(b+a) - (b-a)}{2}$, $a$ is largest when the two even factors are as far apart as possible.
- With product $2560$ fixed, the most lopsided even-by-even split is $2 \times 1280$.
💡 For a fixed product, two numbers are pulled apart most when one of them is as small as the rules allow. Even-only forces the smaller factor down to $2$, not $1$.
8.EE.A.1 Step 4 - Subproblem B: find the units digit of $M = a^2 - 1213$.
- The units digit of $a^2 = 639^2$ depends only on the ones digit of $a$, which is $9$.
- And $9^2 = 81$ ends in $1$.
- Subtract the ones digit of $1213$, which is $3$, working mod $10$.
💡 Units digits only depend on units digits — that is the whole reason mod-$10$ arithmetic exists. No need to compute the full $639^2 = 408{,}321$.
6.EE.A.2 Name the two squares and subtract to remove $M$. Set $M + 1213 = a^2$ and $M + 3 6.NS.B.4 Factor the difference of squares. The right side becomes a product of two intege 4.OA.B.4 Subproblem A: pick the factor pair that maximizes $M$. From $a = \dfrac{(b+a) - 8.EE.A.1 Subproblem B: find the units digit of $M = a^2 - 1213$. The units digit of $a^2 Review
Reasonableness: Spot-check the algebra end-to-end. With $a = 639$, $M = 639^2 - 1213 = 408{,}321 - 1{,}213 = 407{,}108$, whose units digit is indeed $8$. Also check the other condition: $M + 3773 = 407{,}108 + 3{,}773 = 410{,}881 = 641^2$. Both conditions hold and the ones digit lines up with answer (E). Maximality check: any other valid even factor pair of $2560$ — for example $4 \times 640$ — gives $a = 318 < 639$ and a smaller $M$, confirming $2 \times 1280$ is optimal.
Alternative: Tool #3 (Eliminate Possibilities): once $b - a = 2$ and $b + a = 1280$ pin down $a = 639$, you do not need full modular arithmetic to choose among the five units-digit choices. Just compute $639^2$ mentally as $(640-1)^2 = 640^2 - 2 \cdot 640 + 1$, whose ones digit is $0 - 0 + 1 = 1$. Subtract $1213$'s ones digit $3$ to get $1 - 3 \equiv 8$, eliminating (A)-(D). Same answer (E).
CCSS standards used (min grade 8)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Naming the two perfect squares $a^2$ and $b^2$ so that subtracting them eliminates the unknown $M$ and leaves the equation $b^2 - a^2 = 2560$.)6.NS.B.4Find the greatest common factor and least common multiple; apply the distributive property to express a sum as a product (Treating $(b-a)(b+a) = 2560$ as a factor-pair problem and using parity to restrict the pairs to even-by-even splits.)4.OA.B.4Find all factor pairs for a whole number in the range 1-100, and recognize multiples (Choosing the most lopsided even factor pair $2 \times 1280$ of $2560$ to make $a = (b+a - (b-a))/2$ as large as possible.)8.EE.A.1Know and apply the properties of integer exponents (Using the fact that the units digit of $639^2$ depends only on the units digit of $639$, so $639^2 \equiv 9^2 \equiv 1 \pmod{10}$, then subtracting $1213 \equiv 3 \pmod{10}$ to get units digit $8$.)
⭐ When two perfect squares have a fixed difference, factor that difference — the factor pair you pick is exactly the system $b-a = x$, $b+a = y$ that solves for the two square roots. Lopsided pairs give the largest squares.
⭐ When two perfect squares have a fixed difference, factor that difference — the factor pair you pick is exactly the system $b-a = x$, $b+a = y$ that solves for the two square roots. Lopsided pairs give the largest squares.
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