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

The sum of all five numbers is $85$ . Since $85$ is $1$ more than a multiple of $4$ , the number being subtracted must be $1$ more than a multiple of $4$ . Thus, the answer is $\boxed{\textbf{(C)}~17}$ . ~Gavin_Deng

Solution 2

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 answer is $\boxed{\textbf{(C)}~17}$ . ~cxsmi

Solution 3

We try out every number using brute force and get $\boxed{\textbf{(C)}~17}$ .

Note that this is not practical and it is very time-consuming.

Solution 4

Since the sum of $15 + \dots + 19$ is odd, we immediately exclude B and D. We further note that if A is true, then E is true. Hence, the answer is C.

Solution 5

We can use the fact that $(a + b) \mod n = [(a \mod n) + (b \mod n)] \mod n$ . Notice that $15, 16, 17, 18, 19$ dividing by 4 have remainders $3, 0, 1, 2, 3$ . Their sum is 9. It is easy to see that $9-1=8$ is divisible by 4 and so C (17) is the correct answer.

Video Solution 1 by Cool Math Problems

https://youtu.be/BRnILzqVwHk?si=1KwyuFBUDqdMtC6t&t=2

Video Solution 2 by SpreadTheMathLove

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

Video Solution 3

~hsnacademy

Video Solution 4 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution 5 by Daily Dose of Math

~Thesmartgreekmathdude

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

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

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