Difference between revisions of "2020 AIME II Problems/Problem 6"

Latest revision as of 00:03, 25 December 2022

Problem

Define a sequence recursively by $t_1 = 20$ , $t_2 = 21$ , and $t_n = \frac{5t_{n-1}+1}{25t_{n-2}}$ for all $n \ge 3$ . Then $t_{2020}$ can be expressed as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

Let $t_n=\frac{s_n}{5}$ . Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$ . By substitution, we find $s_3 = \frac{53}{50}$ , $s_4=\frac{103}{105\cdot50}$ , $s_5=\frac{101}{105}$ , $s_6=100$ , and $s_7=105$ . So $s_n$ has a period of $5$ . Thus $s_{2020}=s_5=\frac{101}{105}$ . So, $\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}$ . ~mn28407

Solution 2 (Official MAA)

More generally, let the first two terms be $a$ and $b$ and replace $5$ and $25$ in the recursive formula by $k$ and $k^2$ , respectively. Then some algebraic calculation shows that $t_3 = \frac{b\,k+1}{a\, k^2},~~t_4 = \frac{a\, k + b\,k+1}{a\,b\, k^3},~~ t_5 = \frac{a\,k+1}{b\, k^2},~~ t_6 = a, \text{~ and ~}t_7 =b,$ so the sequence is periodic with period $5$ . Therefore $t_{2020} = t_{5} = \frac{20\cdot 5 +1}{21\cdot 25} = \frac{101}{525}.$ The requested sum is $101+525=626$ .

Solution 3

Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$ , so now we are able to determine the numerical value of $t_3$ using this information: $t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500} = \frac{53}{250}$ $t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}$ $t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}$ $t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20$

Alas, we have figured this sequence is period 5! But since $2020 \equiv 5 \pmod 5$ , we can state that $t_{2020} = t_5 = \frac{101}{525}$ . According to the original problem statement, our answer is $\boxed{626}$ . ~ nikenissan

Video Solution

https://youtu.be/_JTWJxbDC1A ~ CNCM

Video Solution 2

https://youtu.be/__B3pJMpfSk

~IceMatrix

Quick way to notice recursion loop

Round the first two values to both be 20. Then, the next element can be rounded to $\frac{1}{5}$ . $t_4$ can then be quickly calculated to around $\frac{1}{250}$ , and $t_5$ can be rounded to $\frac{1}{5}$ . $t_6$ turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally.

@@ Line 27: / Line 27: @@
 ==Video Solution 2==
 https://youtu.be/__B3pJMpfSk
+~IceMatrix
 ==Quick way to notice recursion loop==
-Round the first two values to both be 20. Then, the next element can be rounded to \frac{1}{5}<math>. t_4 can then be quickly calculated to around \frac{1}{250}</math>, and t_5 can be rounded to \frac{1}{5}$. t_6 turns out to be around 20, which means that there is probably a loop with period 5. The rest of the solution proceeds normally.
+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.
-~IceMatrix
 ==See Also==
 {{AIME box|year=2020|n=II|num-b=5|num-a=7}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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