AMC 10 · 2024 · #6

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)

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

Either all positive, or exactly two negatives.

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

2STEP 2

Sweep the positive triples

Closer factors give a smaller sum.

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

The positive minimum

Three times four times five gives the minimum 12.

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

4STEP 4

Set up the negative case

The sum is one positive minus the two sizes.

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

5STEP 5

The negative minimum

Ten with negative six and negative one gives 3.

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

6STEP 6

Compare the two cases

Between 12 and 3 the winner is 3.

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 12 question into ordinary Grade 6-7 work.