2025 AMC 8 Problems/Problem 6

Problem

Sekou writes the numbers $15, 16, 17, 18, 19.$ After he erases one of his numbers, the sum of the remaining four numbers is a multiple of $4.$ Which number did he erase?

$\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 19$

Solution 1

First, we sum the $5$ numbers to get $85$ . 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 $4$ . The sum of the residues of these numbers modulo $4$ is $-1+0+1+2+3=5 \equiv 1 \pmod 4$ . Hence, the number being subtracted must be congruent to $1$ modulo $4$ . The only such number here is $\boxed{\textbf{(C)}~17}$ . ~cxsmi

Solution 3

$15 + 16 + 17 + 18 + 19 = \frac{34}{2} \cdot 5 = 17 \cdot 5 = 85$ , subtracting the first option gives $70$ , the largest mutliple of 4 less or equal to $70$ is $68$ , $85 - 68 = \boxed{\textbf{(C)}~17}$ . ~ alwaysgonnagiveyouup

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

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.

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