2023 AMC 8 Problems/Problem 22
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution (THINKING CREATIVELY!!!)
- 5 Video Solution by Math-X (Smart and Simple)
- 6 Video Solution 1 (Using Diophantine Equations)
- 7 Video Solution 2 by SpreadTheMathLove
- 8 Animated Video Solution
- 9 Video Solution by Magic Square
- 10 Video Solution by Interstigation
- 11 Video Solution by WhyMath
- 12 Video Solution by harungurcan
- 13 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
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 (THINKING CREATIVELY!!!)
~Education, the Study of Everything
Video Solution by Math-X (Smart and Simple)
https://youtu.be/Ku_c1YHnLt0?si=uptT6DExGvKiatZK&t=4952 ~Math-X
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
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.