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

(Video Solution 2 by Thinking Feet)
 
(3 intermediate revisions by 3 users not shown)
Line 1: Line 1:
==Problem==
+
== Problem ==
 +
 
 
Sekou writes the numbers <math>15, 16, 17, 18, 19.</math> After he erases one of his numbers, the sum of the remaining four numbers is a multiple of <math>4.</math> Which number did he erase?
 
Sekou writes the numbers <math>15, 16, 17, 18, 19.</math> After he erases one of his numbers, the sum of the remaining four numbers is a multiple of <math>4.</math> Which number did he erase?
  
 
<math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 19</math>
 
<math>\textbf{(A)}\ 15\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 19</math>
  
==Solution 1==
+
== Solution 1 ==
First, we sum the <math>5</math> numbers to get <math>85</math>. The number subtracted therefore must be 1 more than a multiple of 4. Thus, the answer is <math>\boxed{\textbf{(C)}~17}</math>.
+
 
 +
The sum of all five numbers is <math>85</math>. Since <math>85</math> is <math>1</math> more than a multiple of <math>4</math>, the number subtracted must be <math>1</math> more than a multiple of <math>4</math>. Thus, the answer is <math>\boxed{\textbf{(C)}~17}</math>.
 
~Gavin_Deng
 
~Gavin_Deng
  
==Solution 2==
+
== Solution 2 ==
We consider modulo <math>4</math>. The sum of the residues of these numbers modulo <math>4</math> is <math>-1+0+1+2+3=5 \equiv 1 \pmod 4</math>. Hence, the number being subtracted must be congruent to <math>1</math> modulo <math>4</math>. The only such number here is <math>\boxed{\textbf{(C)}~17}</math>. ~cxsmi
+
The sum of the residues of these numbers modulo <math>4</math> is <math>-1+0+1+2+3=5 \equiv 1 \pmod 4</math>. Hence, the number being subtracted must be congruent to <math>1</math> modulo <math>4</math>. The only such answer is <math>\boxed{\textbf{(C)}~17}</math>.
 +
~cxsmi
  
==Solution 3==
+
== Solution 5 ==  
Since 15 through 19 are all consecutive, the sum of them is 85, which is 1 more than a multiple of 4. Out of all of the solutions, the only one that is a multiple of 4 is (C) 17
 
  
==Solution 4==
+
We try out every number using [[Brute Force]] and get <math>\boxed{\textbf{(C)}~17}</math>
  
We try out every number using Brute Force and get <math>\boxed{\textbf{(C)}~17}</math>
 
 
Note that this is not very practical and it is very time-consuming.
 
Note that this is not very practical and it is very time-consuming.
  
==Video Solution 1 by SpreadTheMathLove==
+
== Video Solution 1 by SpreadTheMathLove ==
 +
 
 
https://www.youtube.com/watch?v=jTTcscvcQmI
 
https://www.youtube.com/watch?v=jTTcscvcQmI
  
==Video Solution by Thinking Feet==
+
==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
 
https://youtu.be/PKMpTS6b988
  
==See Also==
+
==Video Solution by Daily Dose of Math==
 +
 
 +
https://youtu.be/nkpdskFVgdM
 +
 
 +
~Thesmartgreekmathdude
 +
 
 +
== See Also ==
 +
 
 
{{AMC8 box|year=2025|num-b=5|num-a=7}}
 
{{AMC8 box|year=2025|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:Introductory Number Theory Problems]]

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png