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| − | The | + | ==Problem== |
| + | 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> | ||
| + | |||
| + | ==Solution 1== | ||
| + | First, we sum the <math>5</math> numbers to get <math>85</math>. The number subtracted therefore must be 1 more than a multiple of 4. Thus, the answer is <math>\boxed{\textbf{(C)}~17}</math>. | ||
| + | ~Gavin_Deng | ||
| + | |||
| + | ==Solution 2== | ||
| + | We consider modulo <math>4</math>. 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 number here is <math>\boxed{\textbf{(C)}~17}</math>. ~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== | ||
| + | https://youtu.be/PKMpTS6b988 | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2025|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
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. 
