AMC 10 · 2024 · #7

Grade 7 number-theoryarithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

factorsparitysystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: multi-digit-arithmeticprime-factorization

📏 Medium solution 💡 3 insights

Problem

Three integers multiply to 60. Among all such triples, find the smallest sum that is still positive.

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 Identify Subproblems

The sign rule for a positive product splits the search cleanly into two subproblems (Tool #7): Case A — all three factors positive; Case B — one positive and two negative. Each case has only finitely many factor triples of 60, so Tool #2 (Make a Systematic List) finishes the work: list the triples in order, compute each sum, keep the smallest positive value. Comparing the two cases gives the overall minimum.

1STEP 1

Split by sign: a positive product needs an even count of negatives, so all positive or exactly two negative.

Case A: a,b,c ＞ 0. Case B: a ＞ 0, b,c ＜ 0.

2STEP 2

Case A (all positive): list factor triples of 60, pushing factors close together to keep the sum small.

1 · 1 · 60 = 60, sum 62. 1 · 2 · 30 = 60, sum 33. 1 · 3 · 20, 24. 1 · 4 · 15, 20. 1 · 5 · 12, 18. 1 · 6 · 10, 17. 2 · 2 · 15, 19. 2 · 3 · 10, 15. 2 · 5 · 6, 13. 3 · 4 · 5, 12.

3STEP 3

Case A bottoms out at the closest triple 3 + 4 + 5, giving sum 12.

min_Case A(a+b+c) = 3+4+5 = 12.

4STEP 4

Case B: with (p, -q, -r) the sum is p - q - r, positive only when p ＞ q + r — so assign the largest factor to p.

Sum = p - (q+r), need p ＞ q+r for positivity.

5STEP 5

Scan Case B: set p = largest, compute p - q - r, keep positives; the list bottoms out at 10 - 6 - 1 = 3.

min_Case B, positive(p-q-r) = 10 - 6 - 1 = 3, from (10, -6, -1).

6STEP 6

Compare: Case A gives 12, Case B gives 3. The overall least positive sum is 3 — choice (B).

min(12, 3) = 3 → (B).

Answer

Verify the winning triple (10, -6, -1) end to end. Product: 10 · (-6) · (-1) = 10 · 6 = 60, matches. Sum: 10 + (-6) + (-1) = 10 - 7 = 3, positive as required. Could a sum of 1 or 2 exist? Any Case-B triple needs p - q - r ∈ {1, 2}, i.e. p = q + r + 1 or p = q + r + 2, with pqr = 60. Checking divisor triples of 60 exhausts the options: (60,1,1) gives 58, (30,2,1) gives 27, (20,3,1) gives 16, (15,4,1) gives 10, (12,5,1) gives 6, (10,6,1) gives 3, then everything else falls below 3 or below 0. So 3 is truly the minimum and (B) is locked in.

💡Key takeaway

Negative times negative is positive, so two negative factors can cancel — that lets the giant factor 10 shrink down to a sum of just 3. Splitting into sign cases first, then listing factor triples second, turns this AMC 10 question into ordinary Grade 6-7 work.

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