2023 AMC 8 Problems/Problem 22: Difference between revisions
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Solution by Slimeknight | Solution by Slimeknight | ||
=Video Solution by STEM Station(Quick and Easy to Understand):= | |||
https:// | https://youtube.com/watch?v=pNWPAK8wxhk | ||
==Video Solution by Pi Academy== | ==Video Solution by Pi Academy== | ||
Revision as of 20:50, 20 January 2025
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 STEM Station(Quick and Easy to Understand):
https://youtube.com/watch?v=pNWPAK8wxhk
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination