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

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==Solution 1==
 
==Solution 1==
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In this solution, we will use trial and error to solve.
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<math>4000</math> can be expressed as <math>200 \times 20</math>. We divide <math>200</math> by <math>20</math> and get <math>10</math>, divide <math>20</math> by <math>10</math> and get <math>2</math>, and divide <math>10</math> by <math>2</math> to get <math>\boxed{\textbf{(D)}\ 5}</math>. No one said that they have to be in ascending order!
  
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Suppose the first <math>2</math> terms were <math>x</math> and <math>y</math>. Then, the next proceeding terms would be <math>xy</math>, <math>xy^2</math>, <math>x^2y^3</math>, and <math>x^3y^5</math>. Since <math>x^3y^5</math> is the <math>6</math>th term, this must be equal to <math>4000</math>. So, <math>x^3y^5=4000</math>. If we prime factorize <math>4000</math> we get <math>4000 = 5^3 \cdot 2^5</math>. We conclude <math>x=5</math> and <math>y=2</math>, which means that the answer is <math>\boxed{\textbf{(D)}\ 5}</math>
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Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] and clarification edits by apex304
  
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~MrThinker, numerophile (edits apex304)
 
  
 
==Solution 2==
 
==Solution 2==
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In this solution, we will use trial and error to solve.
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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>.
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<math>4000</math> can be expressed as <math>200 \times 20</math>. We divide <math>200</math> by <math>20</math> and get <math>10</math>, divide <math>20</math> by <math>10</math> and get <math>2</math>, and divide <math>10</math> by <math>2</math> to get <math>\boxed{\textbf{(D)}\ 5}</math>. No one said that they have to be in ascending order!
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Solution by [[User:Slimeknight|Slimeknight]]
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==Video Solution by Math-X (Smart and Simple)==
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https://youtu.be/Ku_c1YHnLt0?si=uptT6DExGvKiatZK&t=4952 ~Math-X
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==Video Solution (Solve under 60 seconds!!!)==
 +
https://youtu.be/6O5UXi-Jwv4?si=_Ld6okfFe3jfHzio&t=1020
 +
 
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~hsnacademy
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==Video Solution (THINKING CREATIVELY!!!)==
 +
 
 +
https://youtu.be/LAeSj372-UQ
  
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Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] and clarification edits by apex304
+
~Education, the Study of Everything
  
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==Video Solution 1 by OmegaLearn (Using Diophantine Equations)==
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==Video Solution 1 (Using Diophantine Equations)==
 
https://youtu.be/SwPcIZxp_gY
 
https://youtu.be/SwPcIZxp_gY
  
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==Video Solution by Magic Square==
 
==Video Solution by Magic Square==
 
https://youtu.be/-N46BeEKaCQ?t=2649
 
https://youtu.be/-N46BeEKaCQ?t=2649
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==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=3007
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==Video Solution by WhyMath==
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https://youtu.be/RCYRD7OLSLc
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~savannahsolver
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1249s
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 +
~harungurcan
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==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}}

Latest 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 (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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