Difference between revisions of "2025 AMC 8 Problems/Problem 6"
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== Solution 3 == | == Solution 3 == | ||
| − | We try out every number using [[brute | + | We try out every number using [[brute force]] and get <math>\boxed{\textbf{(C)}~17}</math> |
Note that this is not very practical and it is very time-consuming. | Note that this is not very practical and it is very time-consuming. | ||
Revision as of 00:14, 2 February 2025
Contents
Problem
Sekou writes the numbers 
After he erases one of his numbers, the sum of the remaining four numbers is a multiple of 
Which number did he erase?
Solution 1
The sum of all five numbers is 
. Since is 
more than a multiple of 
, the number subtracted must be more than a multiple of . Thus, the answer is 
.
~Gavin_Deng
Solution 2
The sum of the residues of these numbers modulo is 
. Hence, the number being subtracted must be congruent to modulo . The only such answer is .
~cxsmi
Solution 3
We try out every number using brute force and get
Note that this is not very practical and it is very time-consuming.
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=eMjbNcrUSiSp30ND&t=324 ~hsnacademy
Video Solution 2 by Thinking Feet
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See Also
| 2025 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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. 
