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

A. 15

B. 16

C. 17

D. 18

E. 19

1. 2025 AMC 8 官方解析

Answer (C): The sum of the five numbers is

$15 + 16 + 17 + 18 + 19 = 85.$

Erase one number at a time and compute the sum of the remaining four numbers. For example, if Sekou erases $15$, the remaining sum is $85 - 15 = 70$. The table below lists the possible sums and indicates which sum is a multiple of $4$.

The sum $68$ is the only one that is a multiple of $4$, so Sekou erased the number $17$.

OR

The sum of the five numbers is

$15 + 16 + 17 + 18 + 19 = 85,$

which is $1$ more than $84$, a multiple of $4$. The number $17 = 1 + 16$ is the only number among the five that also is $1$ more than a multiple of $4$. Subtracting $17$ from $85$ gives $68$, which is divisible by $4$. Therefore Sekou erased the number $17$.

OR

The specific numbers in this problem are not important. Consider any five consecutive integers $n$, $n + 1$, $n + 2$, $n + 3$, and $n + 4$. Their sum is $5n + 10$. Erasing the numbers one at a time, the sum of the remaining four numbers will be, respectively, $4n + 10$, $4n + 9$, $4n + 8$, $4n + 7$, and $4n + 6$. Of these numbers, only $4n + 8 = 4(n + 2)$ is divisible by $4$. The number Sekou erased had to be $n + 2$, which is the third of the five numbers.