2024 AMC 10A Problem 7

Since the product is positive, either all three integers are positive or exactly two are negative.

All positive: the sum is at least $3 + 4 + 5 = 12$ (three factors of $60$ cannot all be close to $\sqrt[3]{60} \approx 3.9$ any better than this), far above the small choices.

Two negatives: write the integers as $-x, -y, z$ with $x, y, z > 0$, $xyz = 60$, and sum $z - x - y$. The sum $3$ is achievable: $z = 10$, $xy = 6$, $x + y = 7$ gives $x = 1, y = 6$, i.e. $(-1)(-6)(10) = 60$ with sum $3$.

Since $2$ is an answer choice, check it: we would need $x + y = z - 2$ with $xy = 60/z$. Run through the divisors $z$ of $60$ large enough to matter: $z = 6$: $xy = 10$, need $x+y = 4$ (actual $7$ or $11$); $z = 10$: $xy = 6$, need $8$ (actual $5$ or $7$); $z = 12$: $xy = 5$, need $10$ (actual $6$); $z = 15$: $xy = 4$, need $13$ (actual $4$ or $5$); $z = 20, 30, 60$: $xy = 3, 2, 1$, need $18, 28, 58$ (actual $4, 3, 2$). For $z \le 5$ we would need $x + y \le 3$ while $xy \ge 12$. So sum $2$ is impossible (and sum $1$ fails the same way).

The answer is $\boxed{\textbf{(B)}\ 3}$.

"Integers" (not "positive integers") is the whole point: two negative factors keep the product positive while cancelling most of the sum. Once a small sum is found, the answer choices tell you exactly which smaller values must be ruled out, and a short scan of factor triples does it.

• Assuming all three integers are positive (best sum 3+4+5 = 12, not even a choice), or accepting 3 without ruling out the smaller choice 2.
• Reporting a negative sum such as $(-3)(-4)(5) \to -2$; the problem asks for the least positive sum.