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

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==Problem==
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In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is <math>4000</math>. What is the first term?
  
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<math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10</math>
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==Solution==
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Suppose the first two terms were <math>x</math> and <math>y</math>. Then, the next terms would be <math>xy</math>, <math>xy^2</math>, <math>x^2y^2</math>, and <math>x^3y^5</math>. Since <math>x^3y^5</math> is the sixth term, this must be equal to <math>4000</math>. So, <math>x^3y^5=4000 \Rightarrow (xy)^3y^2=4000</math>. Trying out the choices, we get that <math>x=5</math>, <math>y=2</math>, which means that the answer is $\boxed{\textbf{(D)}\ 5}
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~MrThinker

Revision as of 18:06, 24 January 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

Suppose the first two terms were $x$ and $y$. Then, the next terms would be $xy$, $xy^2$, $x^2y^2$, and $x^3y^5$. Since $x^3y^5$ is the sixth term, this must be equal to $4000$. So, $x^3y^5=4000 \Rightarrow (xy)^3y^2=4000$. Trying out the choices, we get that $x=5$, $y=2$, which means that the answer is $\boxed{\textbf{(D)}\ 5}

~MrThinker

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