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 one very fast and efficient must see==
 
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⠀⣿⣿⠀⠀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⠀⠀
 
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==Solution 2==
+
==Solution 1==
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>.
 
~andliu766
 
  
-rnatog337
+
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.
==Solution 3==
 
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>.
 
  
-SahanWijetunga
+
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.
  
==Solution 4==
+
Therefore, the missing number is <math>\boxed{\textbf{(E) } \text{9}}</math>.
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>.
 
  
~PK-10
+
==Video Solution 1 (Simple and Fast) by Parshwa==
 +
https://youtu.be/Qrf4GTDjjXs
  
==Video Solution 1 (easy to digest) by Power Solve==
+
==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
  
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
https://www.youtube.com/watch?v=9v5q5DxeriM
+
noooooooooooooooooooooooooooooooooooooo!!!!!!!!!!!!!!!!!!
 +
 
 
==Video Solution by Intersigation==
 
==Video Solution by Intersigation==
 
https://youtu.be/ktzijuZtDas&t=232
 
https://youtu.be/ktzijuZtDas&t=232

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 (ProblemsAnswer KeyResources)
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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