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

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==Solution==
 
==Solution==
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The sum of the integers from <math>1</math> to <math>9</math> is <math>45</math>. Since the nearest square number less than <math>45</math> is <math>36</math>, the answer is <math>45 - 36 = \boxed{9}.</math>
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You have to add the sum of the numbers 1 through 9, which is 45. After that, you can use the elimination method with subtraction to each number to determine that the answer is 9.
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==

Revision as of 19:52, 9 October 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

You have to add the sum of the numbers 1 through 9, which is 45. After that, you can use the elimination method with subtraction to each number to determine that the answer is 9.

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=bxVKQiul4AjzVUgC&t=331

~hsnacademy

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🔥]

Video Solution by Intersigation

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

Video Solution by Daily Dose of Math (Understandable, Logical, and Speedy)

https://youtu.be/bSPWqeNO11M?si=HIzlxPjMfvGM5lxR

~Thesmartgreekmathdude

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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