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

(Solution 3)
(Solution 5)
 
(12 intermediate revisions by 8 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 being 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 ==
 +
 
 +
We try out every number using [[brute force]] and get <math>\boxed{\textbf{(C)}~17}</math>.
 +
 
 +
Note that this is not practical and it is very time-consuming.
 +
 
 +
== Solution 4 ==
 +
 
 +
Since the sum of <math>15 + \dots + 19</math> 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 <math>(a + b) \mod n = [(a \mod n) + (b \mod n)] \mod n</math>. Notice that <math>15, 16, 17, 18, 19</math> dividing by 4 have remainders <math>3, 0, 1, 2, 3</math>. Their sum is 9. It is easy to see that <math>9-1=8</math> is divisible by 4 and so C (17) is the correct answer.
 +
 
 +
== Video Solution 1 by Cool Math Problems ==
  
==Solution 3==
+
https://youtu.be/BRnILzqVwHk?si=1KwyuFBUDqdMtC6t&t=2
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
+
 
 +
== Video Solution 2 by SpreadTheMathLove ==
  
==Vide 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 3 ==
 +
 
 +
[//youtu.be/VP7g-s8akMY?si=eMjbNcrUSiSp30ND&t=324 ~hsnacademy]
 +
 
 +
== Video Solution 4 by Thinking Feet ==
 +
 
 
https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
  
==See Also==
+
== Video Solution 5 by Daily Dose of Math ==
 +
 
 +
[//youtu.be/nkpdskFVgdM ~Thesmartgreekmathdude]
 +
==Video Solution(Quick, fast, easy!)==
 +
https://youtu.be/fdG7EDW_7xk
 +
 
 +
~MC
 +
 
 +
== 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 09:25, 16 July 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 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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png