Difference between revisions of "2020 AIME II Problems/Problem 6"
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<cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20</cmath> | <cmath>t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20</cmath> | ||
− | Alas, we have figured this sequence is period 5! | + | Alas, we have figured this sequence is period 5! But since <math>2020 \equiv 5 \pmod 5</math>, we can state that <math>t_{2020} = t_5 = \frac{101}{525}</math>. According to the original problem statement, our answer is <math>\boxed{626}</math>. ~ nikenissan |
==Video Solution== | ==Video Solution== | ||
Line 29: | Line 29: | ||
~IceMatrix | ~IceMatrix | ||
+ | |||
+ | ==Quick way to notice recursion loop== | ||
+ | Round the first two values to both be 20. Then, the next element can be rounded to <math>\frac{1}{5}</math>. <math>t_4</math> can then be quickly calculated to around <math>\frac{1}{250}</math>, and <math>t_5</math> can be rounded to <math>\frac{1}{5}</math>. <math>t_6</math> turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally. | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=5|num-a=7}} | {{AIME box|year=2020|n=II|num-b=5|num-a=7}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:03, 25 December 2022
Contents
Problem
Define a sequence recursively by ,
, and
for all
. Then
can be expressed as
, where
and
are relatively prime positive integers. Find
.
Solution
Let . Then, we have
where
and
. By substitution, we find
,
,
,
, and
. So
has a period of
. Thus
. So,
.
~mn28407
Solution 2 (Official MAA)
More generally, let the first two terms be and
and replace
and
in the recursive formula by
and
, respectively. Then some algebraic calculation shows that
so the sequence is periodic with period
. Therefore
The requested sum is
.
Solution 3
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given and
, so now we are able to determine the numerical value of
using this information:
Alas, we have figured this sequence is period 5! But since , we can state that
. According to the original problem statement, our answer is
. ~ nikenissan
Video Solution
https://youtu.be/_JTWJxbDC1A ~ CNCM
Video Solution 2
~IceMatrix
Quick way to notice recursion loop
Round the first two values to both be 20. Then, the next element can be rounded to .
can then be quickly calculated to around
, and
can be rounded to
.
turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally.
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.