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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>
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==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 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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==Solution 2==
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The sum of the digits from <math>1-9</math> are <math>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45</math>. Note that one of the answer choices (that is equal to Yunju's digits) subtracted from her sum of <math>45</math> must equal a square.
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Since we have all the answer choices, we can check and see which one works. Testing, we have that leaving out <math>9</math> works, so the answer is <math>\boxed{\textbf{(E) }9}</math>.
 
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~andliu766
 
  
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-rnatog337
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Note that <math>6^2 = 36</math> is a very close square to the sum of 45. Checking, we see that <math>45 - 9 = 36 = 6^2</math> works.
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==Solution 3==
 
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Recall from AMC12A 2022 Problem 16, that <math>1+2+\dots+8 = 6^2</math>. Hence removing <math>9</math> works and our answer is <math>\boxed{\textbf{(E) }9}</math>.
 
  
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-SahanWijetunga
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Therefore, the missing number is <math>\boxed{\textbf{(E) } \text{9}}</math>.
  
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==Video Solution 1 (easy to digest) by Power Solve==
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==Video Solution 1 (Simple and Fast) by Parshwa==
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https://youtu.be/Qrf4GTDjjXs
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==Video Solution 2 (easy to digest) by Power Solve==
 
https://youtu.be/HE7JjZQ6xCk?si=sTC7YNSmfEOMe4Sn&t=179
 
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
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 +
~Math-X
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==Video Solution by NiuniuMaths (Easy to understand!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
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https://www.youtube.com/watch?v=9v5q5DxeriM
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noooooooooooooooooooooooooooooooooooooo!!!!!!!!!!!!!!!!!!
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==Video Solution by Intersigation==
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https://youtu.be/ktzijuZtDas&t=232
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=3|num-a=5}}
 
{{AMC8 box|year=2024|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:46, 4 March 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 digits from $1-9$ are $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. Note that one of the answer choices (that is equal to Yunju's digits) subtracted from her sum of $45$ must equal a square.

Note that $6^2 = 36$ is a very close square to the sum of 45. Checking, we see that $45 - 9 = 36 = 6^2$ works.

Therefore, the missing number is $\boxed{\textbf{(E) } \text{9}}$.

Video Solution 1 (Simple and Fast) by Parshwa

https://youtu.be/Qrf4GTDjjXs

Video Solution 2 (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🔥]

noooooooooooooooooooooooooooooooooooooo!!!!!!!!!!!!!!!!!!

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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