Difference between revisions of "2023 AMC 8 Problems/Problem 22"

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Consider the first term is <math>a</math> and the second term is <math>b</math>. Then, the following term will be <math>ab</math>, <math>ab^2</math>, <math>a^2b^3</math> and <math>a^3b^5</math>. Notice that <math>4000=2^5\times 5^3</math>, then we obtain <math>a=\boxed{\textbf{(D)}\ 5}</math> and <math>b=2</math>.
 
Consider the first term is <math>a</math> and the second term is <math>b</math>. Then, the following term will be <math>ab</math>, <math>ab^2</math>, <math>a^2b^3</math> and <math>a^3b^5</math>. Notice that <math>4000=2^5\times 5^3</math>, then we obtain <math>a=\boxed{\textbf{(D)}\ 5}</math> and <math>b=2</math>.
  
Solution by [[User:xana233|xana233]]
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Solution by [[User:Slimeknight|Slimeknight]]
  
 
==Video Solution by Math-X (Smart and Simple)==
 
==Video Solution by Math-X (Smart and Simple)==
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~harungurcan
 
~harungurcan
 +
 +
==Video Solution by Dr. David==
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https://youtu.be/J31l_MwfKT4
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=21|num-a=23}}
 
{{AMC8 box|year=2023|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:36, 10 November 2024

Problem

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$. What is the first term?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$

Solution 1

In this solution, we will use trial and error to solve. $4000$ can be expressed as $200 \times 20$. We divide $200$ by $20$ and get $10$, divide $20$ by $10$ and get $2$, and divide $10$ by $2$ to get $\boxed{\textbf{(D)}\ 5}$. No one said that they have to be in ascending order!

Solution by ILoveMath31415926535 and clarification edits by apex304


Solution 2

Consider the first term is $a$ and the second term is $b$. Then, the following term will be $ab$, $ab^2$, $a^2b^3$ and $a^3b^5$. Notice that $4000=2^5\times 5^3$, then we obtain $a=\boxed{\textbf{(D)}\ 5}$ and $b=2$.

Solution by Slimeknight

Video Solution by Math-X (Smart and Simple)

https://youtu.be/Ku_c1YHnLt0?si=uptT6DExGvKiatZK&t=4952 ~Math-X


Video Solution (Solve under 60 seconds!!!)

https://youtu.be/6O5UXi-Jwv4?si=_Ld6okfFe3jfHzio&t=1020

~hsnacademy

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/LAeSj372-UQ

~Education, the Study of Everything

Video Solution 1 (Using Diophantine Equations)

https://youtu.be/SwPcIZxp_gY

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=ms4agKn7lqc

Animated Video Solution

https://youtu.be/tnv1XzSOagA

~Star League (https://starleague.us)

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=2649

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=3007

Video Solution by WhyMath

https://youtu.be/RCYRD7OLSLc

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1249s

~harungurcan

Video Solution by Dr. David

https://youtu.be/J31l_MwfKT4

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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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