2024 AMC 10B Problem 5

Question

In the following expression, Melanie changed some of the plus signs to minus signs:

1 + 3 + 5 + 7 + \cdots + 97 + 99

When the new expression was evaluated, it was negative. What is the least number of plus signs that Melanie could have changed to minus signs?

14

15

16

17

18

The full sum is

1 + 3 + \cdots + 99 = 50^2 = 2500;

flipping a term

t

lowers the sum by

2t

To go below

0

with the fewest flips, flip the largest terms; their doubled total must exceed

2500

The full sum is

1 + 3 + \cdots + 99 = 50^2 = 2500.

Flipping a term

t

drops the total by

2t,

so to go negative the flipped terms have to add up to more than

1250.

The greedy move is to flip the biggest odd numbers: flipping the top

k

gives

99 + 97 + \cdots = k(100 - k).

We want

k(100 - k) \gt 1250.

At

k = 14

it’s only

1204,

but at

k = 15

it jumps to

1275.

So

15

flips do it. Thus, B is the correct answer.

Problem 5 in other years