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

Revision as of 09:24, 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 23: / Line 23: @@
 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 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$ is divisible by 4 and so C (17) is the correct answer.
 == Video Solution 1 by Cool Math Problems ==

2025 AMC 8 (Problems • Answer Key • Resources)
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