AMC 10 · 2024 · #9
Grade 8 number-theoryalgebraPick an answer.
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 a² and b², subtract to eliminate M, and the unknowns are now an integer pair (a,b) with b² - a² = 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 a² - 1213.
Cancel M
Subtracting leaves a square difference of 2560.
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.EE.A.2Convert To AlgebraFactor it
The two factors must share the same parity.
Difference of squares is the algebra version of arranging b² - a² 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.
A difference of two squares splits into the two numbers added times the two subtracted.
▸ Why?
That factoring is exact, so a size question becomes a question about a product of two whole numbers.
▸ Why?
Once the product is fixed, the possibilities are exactly its factor pairs.
Pick the largest M
A gap of 2 makes the squares largest.
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.
4.OA.B.4Identify SubproblemsRead the units digit
Squaring 639 and subtracting 1213 ends in 8.
Units digits only depend on units digits — that is the whole reason mod-10 arithmetic exists. No need to compute the full 639² = 408,321.
8.EE.A.1Identify SubproblemsWhen 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.