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

 
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==Vide Solution 1 by SpreadTheMathLove==
 
==Vide Solution 1 by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 18:55, 30 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

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

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
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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