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

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 $\boxed{\textbf{(E) }}$

@@ Line 1: / Line 1: @@
 ==Problem==
+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?
+<math>\textbf{(A) } 5\qquad\textbf{(B) } 6\qquad\textbf{(C) } 7\qquad\textbf{(D) } 8\qquad\textbf{(E) } 9</math>
 ==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. 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>

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

Revision as of 14:02, 25 January 2024

Problem

Solution 1