Difference between revisions of "2024 AMC 8 Problems/Problem 4"

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==Problem==
 
==Problem==
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When Yunji added all the integers from <math>1</math> to <math>9</math>, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
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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>
 
==Solution 1==
 
==Solution 1==
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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. However, since this is the AMC 8, we can simply try each option. When <math>x = 9</math>, we see that <math>45 - 9 = 36</math>, a perfect square. Therefore, our answer is <math>\boxed{\textbf{(E) }}</math>

Revision as of 14:02, 25 January 2024

Problem

When Yunji added all the integers from $1$ to $9$, she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?

$\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9$

Solution 1

The sum of the numbers from $1$ to $9$ is \[1 + 2 + 3 + \cdots + 9 = \frac{9(10)}{2} = 45.\] Denote the number left out when adding to be $x$. Thus, $45 - x$ is a perfect square. We could rigorously solve the values of $x$ such that $45 - x$ is a perfect square. However, since this is the AMC 8, we can simply try each option. When $x = 9$, we see that $45 - 9 = 36$, a perfect square. Therefore, our answer is $\boxed{\textbf{(E) }}$