Difference between revisions of "2023 AMC 8 Problems/Problem 22"
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− | <math>4000</math> can be expressed as <math>200 \times 20</math>. We divide <math>200</math> by <math>20</math> and get <math>10</math>, divide <math>20</math> by <math>10</math> and get <math>2</math>, and divide <math>10</math> by <math>2</math> to get | + | <math>4000</math> can be expressed as <math>200 \times 20</math>. We divide <math>200</math> by <math>20</math> and get <math>10</math>, divide <math>20</math> by <math>10</math> and get <math>2</math>, and divide <math>10</math> by <math>2</math> to get <math>\boxed{\textbf{(D)}\ 5}</math>. No one said that they have to be in ascending order! |
Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] | Solution by [[User:ILoveMath31415926535|ILoveMath31415926535]] |
Revision as of 12:08, 25 January 2023
Contents
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
,
, which means that the answer is
~MrThinker, numerophile (edits apex304)
Solution 2
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
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
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.