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

(→‎Solution)
(Solution 2 to Problem 22 on 2023 AMC8)
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~MrThinker
 
~MrThinker
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Solution 2
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We assign A as a term in the sequence. Where <cmath>a_6 = 4000
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We identify by observation and a bit of multiplying for the rest of the terms in the sequence based on the given rules.
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We use C and D as our first 2 numbers
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</cmath>a_1 = C(First Number in the Seqence)
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<cmath>a_2 = D
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...
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</cmath>a_6 = C^3 \cdot D^5 -> 4000<math></math>
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Therefore after prime factorizing 4000 as 5^3 \cdot 2^5,
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We find  <math>\boxed{\text{(D)}5}</math> as our anwser
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~cxrupptedpat, wuwang2002, apex304
  
 
==Animated Video Solution==
 
==Animated Video Solution==

Revision as of 19:54, 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

Solution 2 We assign A as a term in the sequence. Where \[a_6 = 4000 We identify by observation and a bit of multiplying for the rest of the terms in the sequence based on the given rules. We use C and D as our first 2 numbers\]a_1 = C(First Number in the Seqence) \[a_2 = D ...\]a_6 = C^3 \cdot D^5 -> 4000$$ (Error compiling LaTeX. Unknown error_msg) Therefore after prime factorizing 4000 as 5^3 \cdot 2^5, We find $\boxed{\text{(D)}5}$ as our anwser ~cxrupptedpat, wuwang2002, apex304

Animated Video Solution

https://youtu.be/tnv1XzSOagA

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

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