Difference between revisions of "2025 AMC 8 Problems/Problem 6"

Latest revision as of 22:29, 31 January 2025

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 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 5

We try out every number using Brute Force and get $\boxed{\textbf{(C)}~17}$

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

Video Solution 1 by SpreadTheMathLove

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

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=eMjbNcrUSiSp30ND&t=324 ~hsnacademy

Video Solution 2 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution by Daily Dose of Math

https://youtu.be/nkpdskFVgdM

~Thesmartgreekmathdude

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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