Difference between revisions of "2025 AMC 8 Problems/Problem 6"
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==Solution 1== | ==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}. | + | 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 | ~Gavin_Deng | ||
==Solution 2== | ==Solution 2== | ||
| − | We consider modulo < | + | 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 |
Revision as of 20:58, 29 January 2025
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?
A)15 B)16 C)17 D)18 E)19
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
