2024 AMC 8 Problems/Problem 4: Difference between revisions
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<math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math> | <math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math> | ||
==Solution 1== | ==Solution 1== | ||
The sum of the numbers from <math>1</math> to <math>9</math> is <cmath>1 + 2 + 3 + \cdots + 9 = \frac{9(10)}{2} = 45.</cmath> Denote the number left out when adding to be <math>x</math>. Thus, | The sum of the numbers from <math>1</math> to <math>9</math> is <cmath>1 + 2 + 3 + \cdots + 9 = \frac{9(10)}{2} = 45.</cmath> Denote the number left out when adding to be <math>x</math>. Thus, <math>45 - x</math> is a perfect square. We also know that <math>x</math> must be between <math>1</math> and <math>9</math> inclusive, so the answer is <math>\boxed{\textbf{(E) }9}</math>. | ||
-rnatog337 | -rnatog337 | ||
Revision as of 12:35, 26 January 2024
Problem
When Yunji added all the integers from 
to 
, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
Solution 1
The sum of the numbers from to is ![\[1 + 2 + 3 + \cdots + 9 = \frac{9(10)}{2} = 45.\]](http://latex.artofproblemsolving.com/d/2/b/d2bc0143cfd55ae1e08ed5ad87a7236c346290a4.png)
Denote the number left out when adding to be 
. Thus, 
is a perfect square. We also know that must be between and inclusive, so the answer is 
.
-rnatog337
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/HE7JjZQ6xCk?si=sTC7YNSmfEOMe4Sn&t=179
See Also
| 2024 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America. 
