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

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== Solution 1 ==
 
== Solution 1 ==
  
By the formula for the <math>n</math>th term of an arithmetic sequence, we get the answer <math>a+d(n-1)</math> where <math>a=100, d=-7</math> and <math>n=10</math> which is equal to <math>100-7(10-1) = \boxed{\text{(B)\ 37}}</math>.
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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
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== Solution 2 ==
 
== Solution 2 ==
  
Since we want to find the <math>9</math>th number Lucius says after he says <math>100</math>, our answer is <math>100-(9 \cdot 7) = \boxed{\text{(B)\ 37}}</math>
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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>
  
 
~Sigmacuber
 
~Sigmacuber
  
==Solution 3 (Not the most practical)==
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== Solution 3 ==
  
 
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>.
 
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>.
  
Note that this is not the most practical solution, and it is very time-consuming.
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Note that this solution is not practical and very time-consuming.
  
 
~athreyay
 
~athreyay
  
== Video Solution by Pi Academy ==
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==  Video Solution 1 ==
 
 
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
 
 
 
 
 
==  Video Solution 1 (Detailed Explanation) 🚀⚡📊 ==
 
 
 
Youtube Link ⬇️
 
  
 
https://youtu.be/rf5c9ulMA2I
 
https://youtu.be/rf5c9ulMA2I
  
~ ChillGuyDoesMath :)
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~ ChillGuyDoesMath
  
 
== Video Solution by SpreadTheMathLove ==
 
== Video Solution by SpreadTheMathLove ==
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https://youtu.be/PKMpTS6b988
 
https://youtu.be/PKMpTS6b988
  
==Video Solution (A Clever Explanation You’ll Get Instantly)==
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== Video Solution 4 ==
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https://youtu.be/VP7g-s8akMY?si=K8Pxs_TQhlR2ntt9&t=211
 
https://youtu.be/VP7g-s8akMY?si=K8Pxs_TQhlR2ntt9&t=211
 
~hsnacademy
 
~hsnacademy
== Video Solution by CoolMathProblems ==
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== Video Solution 5 by CoolMathProblems ==
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https://youtu.be/nwUanrEZpcQ
 
https://youtu.be/nwUanrEZpcQ
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== Video Solution 6 by Pi Academy ==
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 +
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
  
 
== See Also ==
 
== See Also ==

Revision as of 21:06, 3 February 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.

~athreyay

Video Solution 1

https://youtu.be/rf5c9ulMA2I

~ ChillGuyDoesMath

Video Solution by SpreadTheMathLove

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

Video Solution 2 by Daily Dose of Math

https://youtu.be/rjd0gigUsd0

~Thesmartgreekmathdude

Video Solution 3 by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution 4

https://youtu.be/VP7g-s8akMY?si=K8Pxs_TQhlR2ntt9&t=211 ~hsnacademy

Video Solution 5 by CoolMathProblems

https://youtu.be/nwUanrEZpcQ

Video Solution 6 by Pi Academy

https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK

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

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