2024 AMC 10A Problem 7

The problem says "integers", not "positive integers", so negative numbers are allowed. That matters a lot.

1. Which signs are possible? The product $60$ is positive. So either all three numbers are positive, or exactly two are negative. (One or three negatives would make the product negative.)
2. All positive. Numbers that multiply to $60$ and add to something small should be close together, like $3, 4, 5$, which add to $12$. Every choice from $2$ to $6$ is much smaller than that, so positive numbers are no good.
3. Two negatives. Write the numbers as $-x$, $-y$ and $z$, where $x$, $y$, $z$ are positive, $x \cdot y \cdot z = 60$, and the sum is $z - x - y$. We want $z$ big and $x + y$ just a little smaller than $z$.
4. Try some values of $z$. For each $z$ that divides $60$, the product $x \cdot y$ must be $60 \div z$. Here are the useful cases:

1. For $z = 5$ or smaller, $x \cdot y$ is at least $12$, so $x + y$ is at least $7$ and the sum $z - x - y$ is negative.
2. The smallest positive sum in the table is $3$, from $z = 10$, $x = 1$, $y = 6$: indeed $(-1) \cdot (-6) \cdot 10 = 60$ and $-1 - 6 + 10 = 3$. No row gives $1$ or $2$.

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

Read the word "integers" carefully: it lets two of the numbers be negative, which keeps the product positive while cancelling most of the sum. After you find a small sum, the answer choices tell you exactly which smaller values you must rule out, and a short table of the ways to split $60$ does that.

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 $\boxed{\textbf{(B)}\ 3}$.