Difference between revisions of "2024 AMC 10B Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
| − | + | Recall that the sum of the first <math>n</math> odd numbers is <math>n^2</math>. Thus <cmath>1 + 3 + 5 + 7\dots + 97 + 99 = 50^2 = 2500.</cmath> | |
| + | |||
| + | 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 <cmath>1 + 3 + 5 +\dots + (2n - 3) + (2n - 1) - (2n + 1) - (2n + 3)-\dots - 97 - 99 < 0.</cmath> | ||
| + | |||
| + | The positive section of the sum will contribute <math>n^2</math>, and the negative section will contribute <math>-(2500-n^2) = (n^2 - 2500)</math>. The inequality simplifies to | ||
| + | <cmath>n^2 + (n^2 - 2500) < 0</cmath> | ||
| + | <cmath>2n^2 < 2500</cmath> | ||
| + | <cmath>n^2 < 1250</cmath> | ||
| + | The greatest positive value of <math>n</math> satisfying the inequality is <math>n = 35</math>, corresponding to <math>35</math> positive numbers, and <math>\boxed{\text{B. } 15}</math> negatives. | ||
| + | |||
| + | ~numerophile | ||
==See also== | ==See also== | ||
Revision as of 04:04, 14 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:
![\[1+3+5+7+...+97+99\]](http://latex.artofproblemsolving.com/a/3/0/a307bbc8aeccf32d5dea44474103639a57f21408.png)
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?
Solution 1
Recall that the sum of the first 
odd numbers is 
. Thus ![\[1 + 3 + 5 + 7\dots + 97 + 99 = 50^2 = 2500.\]](http://latex.artofproblemsolving.com/8/2/8/82896f15730b390fa950c78c5f2e90e28ec31d20.png)
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 ![\[1 + 3 + 5 +\dots + (2n - 3) + (2n - 1) - (2n + 1) - (2n + 3)-\dots - 97 - 99 < 0.\]](http://latex.artofproblemsolving.com/9/6/1/961b81f787fed7508b62c89c4db4a4f2830192c0.png)
The positive section of the sum will contribute , and the negative section will contribute 
. The inequality simplifies to
![\[n^2 + (n^2 - 2500) < 0\]](http://latex.artofproblemsolving.com/6/5/5/655fa59f1226d32b5ef977ba01acf5cd1ccfc154.png)
![\[2n^2 < 2500\]](http://latex.artofproblemsolving.com/1/d/4/1d4e05f735e82749bfbfdeae464460bdf690064b.png)
![\[n^2 < 1250\]](http://latex.artofproblemsolving.com/3/2/4/3246341f0a640eda79a8fea455a652a3408e1542.png)
The greatest positive value of satisfying the inequality is 
, corresponding to 
positive numbers, and 
negatives.
~numerophile
See also
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
