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

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

@@ 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?
-A)15 B)16 C)17 D)18 E)19
+<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
-==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 ==
+https://youtu.be/BRnILzqVwHk?si=1KwyuFBUDqdMtC6t&t=2
+== Video Solution 2 by SpreadTheMathLove ==
+https://www.youtube.com/watch?v=jTTcscvcQmI
+== Video Solution 3 ==
+[//youtu.be/VP7g-s8akMY?si=eMjbNcrUSiSp30ND&t=324 ~hsnacademy]
+== Video Solution 4 by Thinking Feet ==
+https://youtu.be/PKMpTS6b988
+== 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 ==
-==Solution 3==
+{{AMC8 box|year=2025|num-b=5|num-a=7}}
-<math>15 + 16 + 17 + 18 + 19 = \frac{34}{2} \cdot 5 = 17 \cdot 5 = 85</math>, subtracting the first option gives <math>70</math>, the largest mutliple of 4 less or equal to <math>70</math> is <math>68</math>, <math>85 - 68 = \boxed{\textbf{(C)}~17}</math>.
+{{MAA Notice}}
-~ alwaysgonnagiveyouup
+[[Category:Introductory Number Theory Problems]]

2025 AMC 8 (Problems • Answer Key • Resources)
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