Find three **consecutive integers** whose sum is $-42$. What is the **smallest** of the three?

Let $n$ be the smallest integer. The three consecutive integers are $n$, $n+1$, $n+2$. Their sum is $-42$: $n + (n+1) + (n+2) = -42$ $3n + 3 = -42$ $3n = -45$ $n = -15$. The three integers are $-15$, $-14$, $-13$. The smallest is $-15$. Check: $-15 + (-14) + (-13) = -42$ ✓. - Trap B ($-14$) is the **middle** of the three — confusion about which the question asks for. - Trap C ($-13$) is the **largest** of the three. - Trap D ($-12$) does not appear in the triple. Shortcut: when three consecutive integers sum to $S$, the **middle** integer is $S/3 = -42/3 = -14$, and the smallest is one less: $-15$.