Difference between revisions of "2023 AMC 8 Problems/Problem 22"
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Consider the first term is <math>a</math> and the second term is <math>b</math>. Then, the following term will be <math>ab</math>, <math>ab^2</math>, <math>a^2b^3</math> and <math>a^3b^5</math>. Notice that <math>4000=2^5\times 5^3</math>, then we obtain <math>a=\boxed{\textbf{(D)}\ 5}</math> and <math>b=2</math>. | Consider the first term is <math>a</math> and the second term is <math>b</math>. Then, the following term will be <math>ab</math>, <math>ab^2</math>, <math>a^2b^3</math> and <math>a^3b^5</math>. Notice that <math>4000=2^5\times 5^3</math>, then we obtain <math>a=\boxed{\textbf{(D)}\ 5}</math> and <math>b=2</math>. | ||
− | Solution by | + | Solution by Slimeknight |
− | == Video | + | ==Video solution by CoolMathProblems: == |
− | https://youtu.be/ | + | https://youtu.be/KcA5CneWAbI |
+ | ==Video Solution by STEM Station(Quick and Easy to Understand)== | ||
+ | https://youtube.com/watch?v=pNWPAK8wxhk | ||
+ | ~STEM Station | ||
− | ==Video Solution by | + | ==Video Solution by Pi Academy== |
− | https://youtu.be/ | + | https://youtu.be/0Fb2lOHTKJo?si=muR8lEE8byZhzvQX |
− | |||
==Video Solution (THINKING CREATIVELY!!!)== | ==Video Solution (THINKING CREATIVELY!!!)== | ||
Line 29: | Line 31: | ||
~Education, the Study of Everything | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
+ | https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2694 | ||
+ | ~hsnacademy | ||
==Video Solution 1 (Using Diophantine Equations)== | ==Video Solution 1 (Using Diophantine Equations)== |
Latest revision as of 15:18, 29 January 2025
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video solution by CoolMathProblems:
- 5 Video Solution by STEM Station(Quick and Easy to Understand)
- 6 Video Solution by Pi Academy
- 7 Video Solution (THINKING CREATIVELY!!!)
- 8 Video Solution (A Clever Explanation You’ll Get Instantly)
- 9 Video Solution 1 (Using Diophantine Equations)
- 10 Video Solution 2 by SpreadTheMathLove
- 11 Animated Video Solution
- 12 Video Solution by Magic Square
- 13 Video Solution by Interstigation
- 14 Video Solution by WhyMath
- 15 Video Solution by harungurcan
- 16 Video Solution by Dr. David
- 17 See Also
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
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
Solution 2
Consider the first term is and the second term is . Then, the following term will be , , and . Notice that , then we obtain and .
Solution by Slimeknight
Video solution by CoolMathProblems:
Video Solution by STEM Station(Quick and Easy to Understand)
https://youtube.com/watch?v=pNWPAK8wxhk ~STEM Station
Video Solution by Pi Academy
https://youtu.be/0Fb2lOHTKJo?si=muR8lEE8byZhzvQX
Video Solution (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=2694 ~hsnacademy
Video Solution 1 (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/DBqko2xATxs&t=3007
Video Solution by WhyMath
~savannahsolver
Video Solution by harungurcan
https://www.youtube.com/watch?v=Ki4tPSGAapU&t=1249s
~harungurcan
Video Solution by Dr. David
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.