AMC 10 · 2024 · #13

Grade 8 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

perfect-squaresprime-factorizationsystematic-enumeration convert-to-algebraidentify-subproblemssystematic-enumeration ↑ Prerequisites: perfect-squaresprime-factorizationlinear-equations-two-var

📏 Medium solution 💡 3 insights

Problem

Positive integers x and y satisfy √(x) + √(y) = √(1183). Find the minimum value of x + y.

Pick an answer.

(A)

585

(B)

595

(C)

623

(D)

700

(E)

791

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

How to solve

Strategy Convert to Algebra

The equation has irrational square roots on both sides, so the standard move is Tool #13 (Convert to Algebra): factor √(1183) to expose its irrational part. Once √(1183) = 13√(7), the only way two positive-integer square roots can sum to 13√(7) is if each one is also of the form (integer)√(7) — that's the subproblem split (Tool #7): first force the form x = 7a², y = 7b², then minimize. Tool #2 (Systematic List) handles the finite minimization: list every positive-integer pair (a, b) with a + b = 13 and pick the one with the smallest a² + b². Classic "sum fixed, minimize sum-of-squares" — closest to equal wins.

1STEP 1

Prime-factor 1183 = 7 · 13², so √(1183) = 13√(7) and the equation becomes √(x) + √(y) = 13√(7).

1183 = 7 · 13² ⟹ √(x) + √(y) = 13√(7)

2STEP 2

Both √(x), √(y) must be multiples of √(7): x = 7a², y = 7b², so a√(7) + b√(7) = 13√(7) gives a + b = 13.

x = 7a², y = 7b² ⟹ a√(7) + b√(7) = 13√(7) ⟹ a + b = 13

3STEP 3

Substitute into the objective: x + y = 7a² + 7b² = 7(a² + b²), so we just minimize a² + b².

x + y = 7a² + 7b² = 7(a² + b²)

4STEP 4

With a + b = 13, a² + b² is smallest when a, b are closest: (6, 7) gives 36 + 49 = 85.

min_a+b=13 (a² + b²) = 6² + 7² = 36 + 49 = 85

5STEP 5

Multiply back: x + y = 7 · 85 = 595. Check: √(252) + √(343) = 6√(7) + 7√(7) = 13√(7) = √(1183).

x + y = 7 · 85 = 595 → (B)

Answer

595

The answer 595 sits squarely in the middle of the choices, which fits a minimum that is non-trivial but not at the extreme. Cross-check: the next pair (a, b) = (5, 8) gives 7 · (25 + 64) = 7 · 89 = 623, exactly choice (C); the pair (4, 9) gives 7 · 97 = 679 (not on the menu), and (3, 10) gives 7 · 109 = 763. The menu (A)=585, (B)=595, (C)=623, (E)=791 matches the actual sum-of-squares ladder very cleanly, with (B) being our minimum and (C) the runner-up — a strong sign the algebra is right.

💡Key takeaway

This AMC 10 problem only needs Grade 8 "irrationals don't mix" plus Grade 4 systematic pair-listing that you already know!

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