2025 AMC 8 Problems/Problem 4

Problem

Lucius is counting backward by $7$ s. His first three numbers are $100$ , $93$ , and $86$ . What is his $10$ th number?

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47$

Solution

By the formula for the $n$ th term of an arithmetic sequence, we get that the answer is $a+d(n-1)$ where $a=100, d=-7$ and $n=10$ which is $100 - 7(10 - 1) = \boxed{\text{(B)\ 37}}$ .

~Soupboy0

Solution 2

To find the solution, we could do 100 - (9 * 7) (because the expression finds 9 terms after) = 100 - 63 = $\boxed{\text{(B)\ 37}}$

~Sigmacuber

Solution 3(Not the most practical)

We could just brute force it and count backward by $7$ . So we would do $100, 93, 86, 79, 72, 65, 58, 51, 44, 37$ . The answer is $\boxed{\text{(B)\ 37}}$

Remember that this is not the most practical solution, and it is very time-consuming.

~codegirl2013

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Daily Dose of Math

https://youtu.be/rjd0gigUsd0

~Thesmartgreekmathdude

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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.

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