Difference between revisions of "2025 AMC 8 Problems/Problem 6"
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| − | ==Problem== | + | == Problem == |
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Sekou writes the numbers <math>15, 16, 17, 18, 19.</math> After he erases one of his numbers, the sum of the remaining four numbers is a multiple of <math>4.</math> Which number did he erase? | Sekou writes the numbers <math>15, 16, 17, 18, 19.</math> After he erases one of his numbers, the sum of the remaining four numbers is a multiple of <math>4.</math> Which number did he erase? | ||
<math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 19</math> | <math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 19</math> | ||
| − | ==Solution 1== | + | == Solution 1 == |
| − | + | ||
| + | The sum of all five numbers is <math>85</math>. Since <math>85</math> is <math>1</math> more than a multiple of <math>4</math>, the number subtracted must be <math>1</math> more than a multiple of <math>4</math>. Thus, the answer is <math>\boxed{\textbf{(C)}~17}</math>. | ||
~Gavin_Deng | ~Gavin_Deng | ||
| − | ==Solution 2== | + | == Solution 2 == |
| − | + | The sum of the residues of these numbers modulo <math>4</math> is <math>-1+0+1+2+3=5 \equiv 1 \pmod 4</math>. Hence, the number being subtracted must be congruent to <math>1</math> modulo <math>4</math>. The only such answer is <math>\boxed{\textbf{(C)}~17}</math>. | |
| + | ~cxsmi | ||
| − | ==Solution | + | == Solution 5 == |
| − | |||
| − | + | 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. | |
| − | + | == Video Solution 1 by SpreadTheMathLove == | |
| − | |||
https://www.youtube.com/watch?v=jTTcscvcQmI | https://www.youtube.com/watch?v=jTTcscvcQmI | ||
| − | ==Video Solution by Thinking Feet== | + | == Video Solution 2 by Thinking Feet == |
| + | |||
https://youtu.be/PKMpTS6b988 | https://youtu.be/PKMpTS6b988 | ||
| − | ==See Also== | + | == See Also == |
| + | |||
{{AMC8 box|year=2025|num-b=5|num-a=7}} | {{AMC8 box|year=2025|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 18:33, 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
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 5
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 2 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. 
