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

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==Solution==
 
==Solution==
  
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 \Rightarrow (xy)^3y^2=4000</math>. Prime factorizing <math>4000</math> we get <math>4000 = 2^5 \times 5^3</math>. We conclude <math>x=5</math>, <math>y=2</math>, which means that the answer is <math>\boxed{\textbf{(D)}\ 5}</math>
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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 \Rightarrow (xy)^3y^2=4000</math>. If we prime factorize <math>4000</math> we get <math>4000 = 2^5 \times 5^3</math>. We conclude <math>x=5</math>, <math>y=2</math>, which means that the answer is <math>\boxed{\textbf{(D)}\ 5}</math>
  
 
~MrThinker, numerophile (edits apex304)
 
~MrThinker, numerophile (edits apex304)

Revision as of 00:33, 25 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 $2$ 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)

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

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