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

(Solution 1)
(Solution 2)
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==Solution 2==
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==wapjgpoaewhgawehgoihewoighoiwhg[owrh[oigawh[oghw[aw==
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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Roses are red
~andliu766
+
violets are blue
 +
Grass is green
 +
Chair
  
-rnatog337
 
 
==Solution 3==
 
==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>.
 
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>.

Revision as of 21:40, 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 one very fast and efficient must see

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣠⣤⣤⣤⣤⣤⣶⣦⣤⣄⡀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣿⡿⠛⠉⠙⠛⠛⠛⠛⠻⢿⣿⣷⣤⡀⠀⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠀⣼⣿⠋⠀⠀⠀⠀⠀⠀⠀⢀⣀⣀⠈⢻⣿⣿⡄⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣸⣿⡏⠀⠀⠀⣠⣶⣾⣿⣿⣿⠿⠿⠿⢿⣿⣿⣿⣄⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⠁⠀⠀⢰⣿⣿⣯⠁⠀⠀⠀⠀⠀⠀⠀⠈⠙⢿⣷⡄⠀ ⠀⠀⣀⣤⣴⣶⣶⣿⡟⠀⠀⠀⢸⣿⣿⣿⣆⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣷⠀ ⠀⢰⣿⡟⠋⠉⣹⣿⡇⠀⠀⠀⠘⣿⣿⣿⣿⣷⣦⣤⣤⣤⣶⣶⣶⣶⣿⣿⣿⠀ ⠀⢸⣿⡇⠀⠀⣿⣿⡇⠀⠀⠀⠀⠹⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⠃⠀ ⠀⣸⣿⡇⠀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠉⠻⠿⣿⣿⣿⣿⡿⠿⠿⠛⢻⣿⡇⠀⠀ ⠀⣿⣿⠁⠀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣧⠀⠀ ⠀⣿⣿⠀⠀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⠀⠀ ⠀⣿⣿⠀⠀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿⠀⠀ ⠀⢿⣿⡆⠀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⡇⠀⠀ ⠀⠸⣿⣧⡀⠀⣿⣿⡇⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⣿⣿⠃⠀⠀ ⠀⠀⠛⢿⣿⣿⣿⣿⣇⠀⠀⠀⠀⠀⣰⣿⣿⣷⣶⣶⣶⣶⠶⠀⢠⣿⣿⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⠀⠀⠀⠀⠀⣿⣿⡇⠀⣽⣿⡏⠁⠀⠀⢸⣿⡇⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⣿⣿⠀⠀⠀⠀⠀⣿⣿⡇⠀⢹⣿⡆⠀⠀⠀⣸⣿⠇⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⢿⣿⣦⣄⣀⣠⣴⣿⣿⠁⠀⠈⠻⣿⣿⣿⣿⡿⠏⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⠀⠀⠈⠛⠻⠿⠿⠿⠿⠋

wapjgpoaewhgawehgoihewoighoiwhg[owrh[oigawh[oghw[aw

Roses are red violets are blue Grass is green Chair

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 (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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