Difference between revisions of "2025 AMC 8 Problems/Problem 6"
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| + | ==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? | 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? | ||
A)15 B)16 C)17 D)18 E)19 | A)15 B)16 C)17 D)18 E)19 | ||
| − | 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}. | ||
| + | ~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}$. | |
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 $\boxed{\textbf{(C)}~17}.
~Gavin_Deng
==Solution 2==
We consider modulo$ (Error compiling LaTeX. Unknown error_msg)4
4
-1+0+1+2+3=5 \equiv 1 \pmod 4
1
4
\boxed{\textbf{(C)}~17}$.
