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Difference between revisions of "2024 AMC 8 Problems/Problem 4"
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==Video Solution by Intersigation==
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https://youtu.be/ktzijuZtDas&t=232
  
 
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==See Also==
 
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Revision as of 02:28, 6 February 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. We can now start to find the closest perfect square to 45. We can immediately see that 36 is the closest, so we can just subtract both numbers to get our final answer of $\boxed{\textbf{(E) }9}$.

~ Nivaar

Solution 2

Since we have all the answer choices, we can check and see which one works. Testing, we have that leaving out $9$ works, so the answer is $\boxed{\textbf{(E) }9}$. ~andliu766

-rnatog337

Solution 3

Recall from AMC12A 2022 Problem 16, that $1+2+\dots+8 = 6^2$. Hence removing $9$ works and our answer is $\boxed{\textbf{(E) }9}$.

-SahanWijetunga

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/HE7JjZQ6xCk?si=sTC7YNSmfEOMe4Sn&t=179

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=9ZUxEGmGam7il9xr&t=907

~Math-X


Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=Ylw-kJkSpq8

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=9v5q5DxeriM

Video Solution by Intersigation

https://youtu.be/ktzijuZtDas&t=232

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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