2019 AMC 10A Problem 5

Nothing says the integers are positive. A run that starts at $-m$ and ends at $m$ sums to $0$, so extending it one more step to $m + 1$ gives a sum of exactly $m + 1$.

Take $m = 44$: the integers $-44, -43, \dots, 44, 45$ sum to $45$. This run has $44 + 1 + 45 = 90$ terms.

Can a longer run work? The sum of $k$ consecutive integers starting at $a$ is $\frac{k(2a + k - 1)}{2}$, so $k(2a + k - 1) = 90$ and $k$ must divide $90$. The largest divisor of $90$ is $90$ itself (with $a = -44$), so $90$ is the maximum. Choice (E), $120$, does not divide $90$.

The answer is $\boxed{\textbf{(D)}\ 90}$.

The sum of $k$ consecutive integers is $k$ times their average, and the average is an integer or half-integer, so $2 \cdot 45 = 90$ must be divisible by $k$. Allowing negatives, you can cancel any symmetric block for free, which is why the longest run is so long. Always ask whether "integers" was secretly "positive integers" in your head.

• Restricting to positive integers and answering 9 (from 1 through 9).
• Picking $120$ because it is the biggest choice, without checking that the number of terms must divide $90$.