Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 8 Problems/Problem 4"
(→‎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. 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>
+
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 13:03, 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 $\boxed{\textbf{(E)}}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_4&oldid=212727"