Difference between revisions of "2023 AMC 8 Problems/Problem 22"
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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</math>. If we prime factorize <math>4000</math> we get <math>4000 = 5^3 \cdot 2^5</math>. We conclude <math>x=5</math> and <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</math>. If we prime factorize <math>4000</math> we get <math>4000 = 5^3 \cdot 2^5</math>. We conclude <math>x=5</math> and <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 10:07, 23 February 2023
Contents
[hide]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 and , which means that the answer is .
~MrThinker, numerophile (edits apex304)
Solution 2
In this solution, we will use trial and error to solve. can be expressed as . We divide by and get , divide by and get , and divide by to get . No one said that they have to be in ascending order!
Solution by ILoveMath31415926535 and clarification edits by apex304
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
Video Solution by Interstigation
https://youtu.be/1bA7fD7Lg54?t=2257
Video Solution by WhyMath
~savannahsolver
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.