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Difference between revisions of "2025 AMC 8 Problems/Problem 4"
 
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==Problem==
 
Lucius is counting backward by <math>7</math>s. His first three numbers are <math>100</math>, <math>93</math>, and <math>86</math>. What is his <math>10</math>th number?
 
Lucius is counting backward by <math>7</math>s. His first three numbers are <math>100</math>, <math>93</math>, and <math>86</math>. What is his <math>10</math>th number?
  
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<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47</math>
  
 
==Solution==
 
==Solution==
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~Soupboy0
 
~Soupboy0
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==Solution 2 (Brute Force)==
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You could brute force(not really recommended) the sequence, and we get: <cmath>100, 93, 86, 79, 72, 65, 58, 51, 44, 37</cmath> Therefore, our answer is <math>\boxed{\text{(B)\ 37}}</math>
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~athreyay
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==Vide Solution 1 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=jTTcscvcQmI
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==Video Solution by Daily Dose of Math==
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https://youtu.be/rjd0gigUsd0
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~Thesmartgreekmathdude
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==Video Solution by Thinking Feet==
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https://youtu.be/PKMpTS6b988
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==See Also==
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{{AMC8 box|year=2025|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 19:16, 30 January 2025

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 (Brute Force)

You could brute force(not really recommended) the sequence, and we get: \[100, 93, 86, 79, 72, 65, 58, 51, 44, 37\] Therefore, our answer is $\boxed{\text{(B)\ 37}}$

~athreyay

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

See Also

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. AMC logo.png

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