2024 AMC 10B Problem 5

Below is the professionally curated solution for Problem 5 of the 2024 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10B solutions, or check the answer key.

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Concepts:summationoptimizationbounding to limit cases

Difficulty rating: 1250

5.

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

1+3+5+7++97+991 + 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?

1414

1515

1616

1717

1818

Solution:

The full sum is 1+3++99=502=2500.1 + 3 + \cdots + 99 = 50^2 = 2500. Flipping a term tt drops the total by 2t,2t, so to go negative the flipped terms have to add up to more than 1250.1250. The greedy move is to flip the biggest odd numbers: flipping the top kk gives 99+97+=k(100k).99 + 97 + \cdots = k(100 - k). We want k(100k)>1250.k(100 - k) \gt 1250. At k=14k = 14 it's only 1204,1204, but at k=15k = 15 it jumps to 1275.1275. So 1515 flips do it. Thus, B is the correct answer.

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