Difference between revisions of "2023 AMC 8 Problems/Problem 22"
Numerophile (talk | contribs) (→Solution) |
Numerophile (talk | contribs) (→Solution) |
||
Line 8: | Line 8: | ||
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^3</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>. 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> | 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^3</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>. 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> | ||
− | ~MrThinker | + | ~MrThinker, numerophile |
==Video Solution 1 by OmegaLearn (Using Diophantine Equations)== | ==Video Solution 1 by OmegaLearn (Using Diophantine Equations)== |
Revision as of 00:19, 25 January 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
Suppose the first two terms were and . Then, the next terms would be , , , and . Since is the sixth term, this must be equal to . So, . Prime factorizing we get . We conclude , , which means that the answer is
~MrThinker, numerophile
Video Solution 1 by OmegaLearn (Using Diophantine Equations)
Animated Video Solution
~Star League (https://starleague.us)
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.