Difference between revisions of "2025 AMC 8 Problems/Problem 4"

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== Problem ==
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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?
  
 
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47</math>
 
<math>\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 37 \qquad \textbf{(C)}\ 42 \qquad \textbf{(D)}\ 44 \qquad \textbf{(E)}\ 47</math>
==Solution==
 
  
By the formula for the <math>n</math>th term of an arithmetic sequence, we get that the answer is <math>a+d(n-1)</math> where <math>a=100, d=-7</math> and <math>n=10</math> which is <math>100 - 7(10 - 1) = \boxed{\text{(B)\ 37}}</math>.
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== Solution 1 ==
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We plug <math>a=100, d=-7</math> and <math>n=10</math> into the formula <math>a+d(n-1)</math> for the <math>n</math>th term of an arithmetic sequence whose first term is <math>a</math> and common difference is <math>d</math> to get <math>100-7(10-1) = \boxed{\text{(B)\ 37}}</math>.
  
 
~Soupboy0
 
~Soupboy0
  
==Solution 2 (Brute Force)==
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== Solution 2 ==
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Since we want to find the <math>9</math>th number Lucius says after he says <math>100</math>, <math>7</math> is subtracted from his number <math>9</math> times, so our answer is <math>100-(9 \cdot 7) = \boxed{\text{(B)\ 37}}</math>
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~Sigmacuber
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== Solution 3 ==
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Using [[brute force]] and counting backward by <math>7</math>s, we have <math>100, 93, 86, 79, 72, 65, 58, 51, 44, \boxed{\text{(B)\ 37}}</math>.
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Note that this solution is not practical and very time-consuming.
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~codegirl2013, athreyay
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== Solution 4 ==
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This can be thought of as an arithmetic sequence. Knowing that our first term is <math>100</math>, we have to add <math>7</math> to get to our  0th term, <math>107</math>. Our answer is then <math>107 - 10 \cdot 7 = \boxed{\text{(B)\ 37}}</math>.
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~Kapurnicus, NYCnerd
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== Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
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https://www.youtube.com/watch?v=rf5c9ulMA2I
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~ ChillThingz :)
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== Video Solution 2 ==
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[//www.youtube.com/jTTcscvcQmI SpreadTheMathLove]
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== Video Solution 3 by Daily Dose of Math ==
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[//youtu.be/rjd0gigUsd0 ~Thesmartgreekmathdude]
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== Video Solution 4 ==
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[//youtu.be/PKMpTS6b988 Thinking Feet]
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== Video Solution 5 ==
  
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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[//youtu.be/VP7g-s8akMY?si=K8Pxs_TQhlR2ntt9&t=211 ~hsnacademy]
  
~athreyay
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== Video Solution 6 ==
  
==Vide Solution 1 by SpreadTheMathLove==
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[//youtu.be/nwUanrEZpcQ CoolMathProblems]
https://www.youtube.com/watch?v=jTTcscvcQmI
 
  
==Video Solution by Daily Dose of Math==
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== Video Solution 7 ==
  
https://youtu.be/rjd0gigUsd0
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[//youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK Pi Academy]
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==Video Solution(Quick, fast, easy!)==
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https://youtu.be/fdG7EDW_7xk
  
~Thesmartgreekmathdude
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~MC
  
==Video Solution by Thinking Feet==
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== See Also ==
https://youtu.be/PKMpTS6b988
 
  
==See Also==
 
 
{{AMC8 box|year=2025|num-b=3|num-a=5}}
 
{{AMC8 box|year=2025|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 11:45, 7 September 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 1

We plug $a=100, d=-7$ and $n=10$ into the formula $a+d(n-1)$ for the $n$th term of an arithmetic sequence whose first term is $a$ and common difference is $d$ to get $100-7(10-1) = \boxed{\text{(B)\ 37}}$.

~Soupboy0

Solution 2

Since we want to find the $9$th number Lucius says after he says $100$, $7$ is subtracted from his number $9$ times, so our answer is $100-(9 \cdot 7) = \boxed{\text{(B)\ 37}}$

~Sigmacuber

Solution 3

Using brute force and counting backward by $7$s, we have $100, 93, 86, 79, 72, 65, 58, 51, 44, \boxed{\text{(B)\ 37}}$.

Note that this solution is not practical and very time-consuming.

~codegirl2013, athreyay

Solution 4

This can be thought of as an arithmetic sequence. Knowing that our first term is $100$, we have to add $7$ to get to our 0th term, $107$. Our answer is then $107 - 10 \cdot 7 = \boxed{\text{(B)\ 37}}$.

~Kapurnicus, NYCnerd

Video Solution 1 (Detailed Explanation) 🚀⚡📊

https://www.youtube.com/watch?v=rf5c9ulMA2I

~ ChillThingz :)

Video Solution 2

SpreadTheMathLove

Video Solution 3 by Daily Dose of Math

~Thesmartgreekmathdude

Video Solution 4

Thinking Feet

Video Solution 5

~hsnacademy

Video Solution 6

CoolMathProblems

Video Solution 7

Pi Academy

Video Solution(Quick, fast, easy!)

https://youtu.be/fdG7EDW_7xk

~MC

See Also

2025 AMC 8 (ProblemsAnswer KeyResources)
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

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