2024 AMC 10B Problem 5

There are $50$ odd numbers from $1$ to $99$, and their sum is $50^2 = 2500$.

Changing the sign of a term $t$ lowers the total by $2t$. If the flipped terms have sum $S$, the new value is $2500 - 2S$, which is negative exactly when $S > 1250$.

To reach $S > 1250$ with as few terms as possible, flip the largest terms. The $k$ largest odd numbers are $99, 97, \dots, 101 - 2k$, an arithmetic series with average $100 - k$, so their sum is $k(100 - k)$.

• $k = 14$: $14 \cdot 86 = 1204 < 1250$, not enough.
• $k = 15$: $15 \cdot 85 = 1275 > 1250$, works.

The answer is $\boxed{\textbf{(B)}\ 15}$.

Sign changes are best viewed as subtracting double the term from a known total, which turns "make it negative" into a clean threshold: the flipped terms must carry more than half the total. A greedy choice (take the biggest terms) minimizes the count for any threshold. The closed form $k(100 - k)$ makes checking the boundary instant.

• Flipping terms until the negatives exceed the positives but forgetting that the flipped terms were removed from the positive side too, i.e. comparing to 2500 instead of 1250.
• Stopping at $k = 14$ because $1204$ is "close to" $1250$, or requiring $S \ge 1250$ and not checking that $1250$ itself is unattainable.