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.