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2024 AMC 10B Problem 5

Problem 5 of 25EasierAlgebra

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?

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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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Concepts: summation · optimization · bounding to limit cases

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.