2024 AMC 10B Problems/Problem 5

The following problem is from both the 2024 AMC 10B #5 and 2024 AMC 12B #5, so both problems redirect to this page.

Problem

In the following expression, Melanie changed some of the plus signs to minus signs: $1+3+5+7+...+97+99$ 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?

$\textbf{(A) } 14 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 16 \qquad\textbf{(D) } 17 \qquad\textbf{(E) } 18$

Solution 1

Recall that the sum of the first $n$ odd numbers is $n^2$ . Thus $1 + 3 + 5 + 7+ \dots + 97 + 99 = 50^2 = 2500.$

If we want to minimize the number of sign flips to make the number negative, we must flip the signs corresponding to the largest absolute value. This will result in the inequality $1 + 3 + 5 +\dots + (2n - 3) + (2n - 1) - (2n + 1) - (2n + 3)-\dots - 97 - 99 < 0.$

The positive section of the sum will contribute $n^2$ , and the negative section will contribute $-(2500-n^2) = (n^2 - 2500)$ . The inequality simplifies to $n^2 + (n^2 - 2500) < 0$ $2n^2 < 2500$ $n^2 < 1250$ The greatest positive value of $n$ satisfying the inequality is $n = 35$ , corresponding to $35$ positive numbers, and $\boxed{\text{B. } 15}$ negatives.

~numerophile

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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2024 AMC 10B Problems/Problem 5

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Solution 1

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