Difference between revisions of "2024 AMC 8 Problems/Problem 4"
(→Video Solution 1(easy to digest) by Power Solve) |
Countmath1 (talk | contribs) m (→Solution 1) |
||
| Line 4: | Line 4: | ||
<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, <math>45 - x</math> is a perfect square. We could rigorously solve the values of <math>x</math> such that <math>45 - x</math> is a perfect square. | + | 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 could rigorously solve the values of <math>x</math> such that <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 13:33, 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 could rigorously solve the values of such that is a perfect square. We also know that must be between and inclusive, so the answer is 
.
-rnatog337
