2023 AMC 8 Problems/Problem 22

Revision as of 00:32, 25 January 2023 by Apex304 (talk | contribs) (Solution)

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

Suppose the first $2$ terms were $x$ and $y$. Then, the next terms would be $xy$, $xy^2$, $x^2y^3$, and $x^3y^5$. Since $x^3y^5$ is the sixth term, this must be equal to $4000$. So, $x^3y^5=4000 \Rightarrow (xy)^3y^2=4000$. Prime factorizing $4000$ 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)

Video Solution 1 by OmegaLearn (Using Diophantine Equations)

https://youtu.be/SwPcIZxp_gY

Animated Video Solution

https://youtu.be/tnv1XzSOagA

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


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

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