2019 AMC 10A Problem 5

The problem says "integers", not "positive integers". Negative numbers are allowed, and they can cancel positive ones.

Part 1: a run of 90 integers works.

1. Look at a small example first: $-2, -1, 0, 1, 2$ adds to $0$, because each number cancels its opposite. Put $3$ on the end and the sum becomes $3$.
2. Do the same with $45$: the run $-44, -43, \ldots, 43, 44$ adds to $0$. Put $45$ on the end and the sum is $45$.
3. Count the run $-44, \ldots, 45$: there are $44$ negative numbers, then $0$, then $45$ positive numbers, so $44 + 1 + 45 = 90$ integers.

Part 2: no longer run works.

1. For $k$ consecutive integers, the average is the middle value. It is either a whole number (when $k$ is odd) or a number ending in $.5$ (when $k$ is even).
2. The sum is $k \times (\text{average})$, so $2 \times \text{sum} = k \times (2 \times \text{average})$, and $2 \times \text{average}$ is a whole number.
3. So $k$ must divide $2 \times 45 = 90$. No number bigger than $90$ divides $90$, so $k$ cannot be more than $90$. (This rules out choice (E), $120$.)

The run of $90$ works and nothing longer can, so the greatest number is $90$.

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

Adding a matching negative for every positive number costs nothing, so a run can be stretched much further than you might expect. The limit comes from the average: $k$ consecutive integers always have a sum equal to $k$ times a whole or half-whole number, so $k$ must divide twice the sum. And always check whether "integers" quietly became "positive integers" in your head.