Difference between revisions of "2024 AMC 8 Problems/Problem 4"
(→Solution 2) |
|||
Line 11: | Line 11: | ||
-rnatog337 | -rnatog337 | ||
+ | ==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 | ||
==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== |
Revision as of 20:42, 26 January 2024
Contents
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, so the answer is .
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
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/HE7JjZQ6xCk?si=sTC7YNSmfEOMe4Sn&t=179
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
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.