2023 AMC 8 Problems/Problem 22

Revision as of 11:08, 25 January 2023 by Ilovemath31415926535 (talk | contribs) (Solution 2)

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, $x^3y^5=4000 \Rightarrow (xy)^3y^2=4000$. If we prime factorize $4000$ we get $4000 = 2^5 \times 5^3$. We conclude $x=5$, $y=2$, which means that the answer is $\boxed{\textbf{(D)}\ 5}$

~MrThinker, numerophile (edits apex304)

Solution 2

$4000$ 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

Video Solution 1 by OmegaLearn (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

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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