2024 AMC 10B Problem 5

1. First find the total with all plus signs. Sums of odd numbers starting at $1$ make perfect squares: $1 = 1$, $1 + 3 = 4$, $1 + 3 + 5 = 9$. There are $50$ odd numbers from $1$ to $99$, so the total is $50^2 = 2500$.
2. See what one sign change does. If $99$ becomes $-99$, the total loses $99$ (it is no longer added) and loses another $99$ (it is now subtracted). So the total drops by $2 \cdot 99$.
3. So if the numbers Melanie flips add up to $S$, the new total is $2500 - 2S$. This is negative exactly when $2S > 2500$, that is, when $S > 1250$.
4. To pass $1250$ with as few numbers as possible, flip the biggest ones: $99, 97, 95, \ldots$
5. Add up the $14$ biggest, $99$ down to $73$. They pair up as $99 + 73 = 172$, $97 + 75 = 172$, and so on, giving $7$ pairs:
$$7 \cdot 172 = 1204.$$
This is less than $1250$, so $14$ flips are not enough.
6. Add the next one, $71$: $1204 + 71 = 1275$, which is more than $1250$. So $15$ flips work.

Since even the $14$ biggest numbers fall short, no set of $14$ or fewer can work, and $15$ is the least.

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

Changing a plus to a minus takes away twice that number, so "make the total negative" becomes "the flipped numbers must add up to more than half the total." To use as few flips as possible, always take the biggest numbers first. If the biggest $14$ cannot reach the target, no other $14$ can either.

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 $\boxed{\textbf{(B)}\ 15}$.