2025 AMC 8 Problems/Problem 6

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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