Difference between revisions of "2024 AMC 10B Problems/Problem 5"

Revision as of 12:02, 15 November 2024

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 values with 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

Solution 2

The formula for the sum of all odd positive integers from $1$ to $2n-1$ is $1+3+5+...+2n-3+2n-1=n^2.$ Therefore, the given sum evaluates to $99=2(50)-1\implies 50^2=2500.$ Since we're looking for the minimum possible sign changes, we focus on the largest numbers in the set to change to negative.

If we change the sign of a number $n$ to negative, then the sum decreases by $2n$ . Therefore, we're looking for a subset of numbers that add to greater than $2500/2=1250$ .

Now we look at the answer choices.

$\text{(A) }14$ means that we're changing the signs of the numbers $\{73,75,...,99\}$ , and $73+75+...+99=\dfrac{(73+99)\cdot(\frac{99-73}{2}+1)}{2}=1204$ .

Now this so slightly happens to be less than $1250$ , and $\text{(B) } 15$ means we're adding $71$ to the set, too. Since $71>1250-1204=46$ , then the answer is $\boxed{\text{(B) }15}$ ~Tacos_are_yummy_1

Solution 3

Note that as mentioned above, $1 + 3 + 5... + 2n - 1 = n^{2}$ . So, the sum of the first 50 odd numbers is equal to 2500, and we need to find the smallest integer such that $2k^{2} < 50^{2}$ . Taking the square root, $k = \floor{25\sqrt{2}}$ (Error compiling LaTeX. Unknown error_msg), keeping in mind we need to take the floor $2k^{2}$ needs to be slightly less than 2500. Using an approximation of $\sqrt{2} = 1.4, 25 * \frac{7}{5}$ is 35, so our answer is $\boxed{\text{(B) }15}$ ~aleyang

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/DIl3rLQQkQQ?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

@@ Line 32: / Line 32: @@
 Now this so slightly happens to be less than <math>1250</math>, and <math>\text{(B) } 15</math> means we're adding <math>71</math> to the set, too. Since <math>71>1250-1204=46</math>, then the answer is <math>\boxed{\text{(B) }15}</math> ~Tacos_are_yummy_1
+==Solution 3==
+Note that as mentioned above, <math>1 + 3 + 5... + 2n - 1 = n^{2}</math>. So, the sum of the first 50 odd numbers is equal to 2500, and we need to find the smallest integer such that <math>2k^{2} < 50^{2}</math>. Taking the square root, <math>k = \floor{25\sqrt{2}}</math>, keeping in mind we need to take the floor <math>2k^{2}</math> needs to be slightly less than 2500. Using an approximation of <math>\sqrt{2} = 1.4, 25 * \frac{7}{5}</math> is 35, so our answer is <math>\boxed{\text{(B) }15}</math>
+~aleyang
 ==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

2024 AMC 10B (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 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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