Difference between revisions of "2023 AMC 8 Problems/Problem 22"
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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> | <math>\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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> | 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) | ||
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| + | ==Solution 2== | ||
| + | |||
| + | <math>4000</math> can be expressed as <math>200 \times 20</math>. 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! | ||
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| + | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] | ||
==Video Solution 1 by OmegaLearn (Using Diophantine Equations)== | ==Video Solution 1 by OmegaLearn (Using Diophantine Equations)== | ||
Revision as of 11:06, 25 January 2023
Contents
Problem
In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is 
. What is the first term?
Solution 1
Suppose the first 
terms were 
and 
. Then, the next proceeding terms would be 
, 
, 
, and 
. Since is the 
th term, this must be equal to . So, 
. If we prime factorize we get 
. We conclude 
, 
, which means that the answer is 
~MrThinker, numerophile (edits apex304)
Solution 2
can be expressed as 
. 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)
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=ms4agKn7lqc
Animated Video Solution
~Star League (https://starleague.us)
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=2649
See Also
| 2023 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
