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

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-SahanWijetunga
 
-SahanWijetunga
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==Solution 4==
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We can add all the numbers from 1 through 9 using Gauss's trick, and find that the answer is 45. Since we know that all the square numbers below 45 are 36, 25, 16, 9, 4, 1, and 0, and 36 is the only square number that is within 10 numbers from 45, we can conclude that the answer is 45-36, hence the answer is <math>\boxed{\textbf{(E) }9}</math>.
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-PK-10
  
 
==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==

Revision as of 00:17, 7 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

Solution 4

We can add all the numbers from 1 through 9 using Gauss's trick, and find that the answer is 45. Since we know that all the square numbers below 45 are 36, 25, 16, 9, 4, 1, and 0, and 36 is the only square number that is within 10 numbers from 45, we can conclude that the answer is 45-36, hence the answer is $\boxed{\textbf{(E) }9}$.

-PK-10

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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