Difference between revisions of "2025 AMC 8 Problems/Problem 6"
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==Solution 3== | ==Solution 3== | ||
| − | + | Since 15 through 19 are all consecutive, the sum of them is 85, which is 1 more than a multiple of 4. Out of all of the solutions, the only one that is a multiple of 4 is (C) 17 | |
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==Vide Solution 1 by SpreadTheMathLove== | ==Vide Solution 1 by SpreadTheMathLove== | ||
Latest revision as of 00:15, 31 January 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
First, we sum the 
numbers to get 
. The number subtracted therefore must be 1 more than a multiple of 4. Thus, the answer is 
.
~Gavin_Deng
Solution 2
We consider modulo 
. The sum of the residues of these numbers modulo is 
. Hence, the number being subtracted must be congruent to 
modulo . The only such number here is . ~cxsmi
Solution 3
Since 15 through 19 are all consecutive, the sum of them is 85, which is 1 more than a multiple of 4. Out of all of the solutions, the only one that is a multiple of 4 is (C) 17
Vide Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution by Thinking Feet
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. 
