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

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==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
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==Video Solution by WhyMath==
 
==Video Solution by WhyMath==

Revision as of 21:42, 23 November 2023

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

Suppose the first $2$ terms were $x$ and $y$. Then, the next proceeding terms would be $xy$, $xy^2$, $x^2y^3$, and $x^3y^5$. Since $x^3y^5$ is the $6$th term, this must be equal to $4000$. So, $x^3y^5=4000$. If we prime factorize $4000$ we get $4000 = 5^3 \cdot 2^5$. We conclude $x=5$ and $y=2$, which means that the answer is $\boxed{\textbf{(D)}\ 5}$.

~MrThinker, numerophile (edits apex304)

Solution 2

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

Video Solution (THINKING CREATIVELY!!!)

https://youtu.be/LAeSj372-UQ

~Education, the Study of Everything

Video Solution by Math-X (Smart and Simple)

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

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

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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