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

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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==
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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. 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>.
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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 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 13:35, 26 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 also know that $x$ must be between $1$ and $9$ inclusive, so the answer is $\boxed{\textbf{(E) }9}$.

-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

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