Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎Solution 1)
(36 intermediate revisions by 11 users not shown)
Line 3: Line 3:
  
 
<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 1==
 
==Solution 1==
−
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. 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 <math>\boxed{\textbf{(E) }9}</math>.
 
  
−
~ Nivaar
+
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.
 +
 
 +
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.
 +
 
 +
Therefore, the missing number is <math>\boxed{\textbf{(E) } \text{9}}</math>.
  
−
==Solution 2==
+
==Video Solution 1 (Simple and Fast) by Parshwa==
−
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>.
+
https://youtu.be/Qrf4GTDjjXs
−
~andliu766
 
  
−
-rnatog337
+
==Video Solution 2 (easy to digest) by Power Solve==
−
==Solution 3==
+
https://youtu.be/HE7JjZQ6xCk?si=sTC7YNSmfEOMe4Sn&t=179
−
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
 
 
==Video Solution by Math-X (First fully understand the problem!!!)==
 
==Video Solution by Math-X (First fully understand the problem!!!)==
 
https://youtu.be/BaE00H2SHQM?si=9ZUxEGmGam7il9xr&t=907
 
https://youtu.be/BaE00H2SHQM?si=9ZUxEGmGam7il9xr&t=907
Line 22: Line 23:
 
~Math-X
 
~Math-X
  
−
==Video Solution 1 (easy to digest) by Power Solve==
 
−
https://youtu.be/HE7JjZQ6xCk?si=sTC7YNSmfEOMe4Sn&t=179
 
  
 
==Video Solution by NiuniuMaths (Easy to understand!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
Line 35: Line 34:
 
== 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==
 +
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}}

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_4&oldid=216470"