Difference between revisions of "2024 AMC 8 Problems/Problem 4"
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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>. | 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 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== |
Revision as of 23:18, 6 February 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Video Solution 1 (easy to digest) by Power Solve
- 7 Video Solution by Math-X (First fully understand the problem!!!)
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution 2 by SpreadTheMathLove
- 10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 11 Video Solution by Intersigation
- 12 See Also
Problem
When Yunji added all the integers from to , she mistakenly left out a number. Her incorrect sum turned out to be a square number. What number did Yunji leave out?
Solution 1
The sum of the numbers from to is Denote the number left out when adding to be . Thus, is a perfect square. We also know that must be between and 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 .
~ Nivaar
Solution 2
Since we have all the answer choices, we can check and see which one works. Testing, we have that leaving out works, so the answer is . ~andliu766
-rnatog337
Solution 3
Recall from AMC12A 2022 Problem 16, that . Hence removing works and our answer is .
-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 .
~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 (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.