AMC 10 · 2024 · #15

Grade 8 number-theoryalgebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

perfect-squaresfactorsunits-digit-trackingsystematic-enumeration convert-to-algebraunits-digit-trackingcasework ↑ Prerequisites: perfect-squaresprime-factorizationmulti-digit-arithmetic

📏 Short solution 💡 3 insights

Problem

Find the largest integer M such that M + 1213 and M + 3773 are both perfect squares. Then report the units digit of M.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

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.

1STEP 1

Name the two squares M + 1213 = a² and M + 3773 = b², subtract to cancel M, leaving a fixed gap of 2560.

b² - a² = (M+3773) - (M+1213) = 2560

2STEP 2

Factor it as (b-a)(b+a) = 2560; same parity plus an even product means both must be even.

(b-a)(b+a) = 2560, b-a ≡ b+a (mod 2)

3STEP 3

Pushing the even factors farthest apart gives 2 × 1280, so b = 641 and a = 639.

b - a = 2, b + a = 1280 → b = 641, a = 639

4STEP 4

Units digits only need units digits: 9² ends in 1, minus the 3 from 1213 gives units digit 8.

M ≡ a² - 1213 ≡ 9² - 3 ≡ 1 - 3 ≡ -2 ≡ 8 (mod 10) → (E)

Answer

Spot-check the algebra end-to-end. With a = 639, M = 639² - 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 = 641². 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 × 640 — gives a = 318 ＜ 639 and a smaller M, confirming 2 × 1280 is optimal.

💡Key takeaway

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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