2021 AIME I Problems/Problem 10


Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$, if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$, then

\[a_{k+1} = \frac{m + 18}{n+19}.\]Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$.

Solution 1

We know that $a_{1}=\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$. Note also that $a_{2}=\tfrac{2038}{2040}=\tfrac{1019}{1020}=\tfrac{t}{t+1}$ for $t=1019$. Then $a_{2+q}=\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ (by the Euclidean Algorithm), or $q+1$ divides $1001$. Thus, the least value of $q$ is $6$ and $j=2+6=8$. We know $a_{8}=\tfrac{1019+108}{1020+114}=\tfrac{1127}{1134}=\tfrac{161}{162}$. Now $a_{8+q}=\tfrac{161+18q}{162+19q}$ unless $18q+161$ and $19q+162$ are not relatively prime which happens the first time $q+1$ divides $18q+161$ or $q+1$ divides $143$ or $q=10$, and $j=8+10=18$. We have $a_{18}=\tfrac{161+180}{162+190}=\tfrac{341}{352}=\tfrac{31}{32}$. Now $a_{18+q}=\tfrac{31+18q}{32+19q}$ unless $18q+31$ and $19q+32$ are not relatively prime. This happens the first time $q+1$ divides $18q+31$ implying $q+1$ divides $13$, which is prime so $q=12$ and $j=18+12=30$. We have $a_{30}=\tfrac{31+216}{32+228}=\tfrac{247}{260}=\tfrac{19}{20}$. We have $a_{30+q}=\tfrac{18q+19}{19q+20}$, which is always reduced by EA, so the sum of all $j$ is $1+2+8+18+30=\boxed{059}$.



Whenever a fraction is in the form $\frac{t}{t+1}$ in lowest terms, the difference between the numerator and the denominator in the original fraction will always divide the numerator. We can model $a_j$ as $\frac{m+18k}{m+19k+1}$ (not necessarily simplified) if $a_{j-k}=\frac{m}{m+1}$ for integers $j$ and $k$. We want the least $k$ such that $\gcd(k+1,{m+18k})\neq1$. Let $d$ be a divisor of both $k+1$ and $m+18k$, then $d\mid18k+18$, so $d\mid{m-18}$. This follows from the Euclidean Algorithm.


Solution 2 (Euclidean Algorithm and Generalization)

Let $a_{j_1}, a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ be all terms in the form $\frac{t}{t+1},$ where $j_1<j_2<j_3<\cdots<j_u,$ and $t$ is some positive integer.

We wish to find $\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\frac{m}{m+1}$ for some positive integer $m.$

To find $\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\boldsymbol{k'}$ for which \[\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}}\] is reducible:

If $\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm, we have \begin{align*} d\mid m+18k' \text{ and } d\mid m+1+19k' &\implies d\mid m+18k' \text{ and } d\mid k'+1 \\ &\implies d\mid m-18 \text{ and } d\mid k'+1. \end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$

We will prove that for such value of $\boldsymbol{k',}$ the number $\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\boldsymbol{\frac{t}{t+1}:}$ \[a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}=\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\frac{\frac{m-18}{k'+1}+18}{\frac{m-18}{k'+1}+19}, \hspace{10mm} (*)\] where $t=\frac{m-18}{k'+1}+18$ must be a positive integer.

We start with $m=2020$ and $a_{j_1}=a_1=\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ by filling out the table below recursively: \[\begin{array}{c|c|c|c|c|c}  & & & & & \\ [-2ex] \boldsymbol{i} & \boldsymbol{m} & \boldsymbol{m-18} & \boldsymbol{k'+1} & \boldsymbol{k'} & \boldsymbol{a_{j_{i+1}} \left(\textbf{by } (*)\right)} \\ [0.5ex]  \hline   & & & & & \\ [-1.5ex]  1 & 2020 & 2002 & 2 & 1 & \hspace{4.25mm} a_2 = \frac{1019}{1020} \\ [1ex]      2 & 1019 & 1001 & 7 & 6 & \hspace{2.75mm} a_8 = \frac{161}{162} \\ [1ex]     3 & 161 & 143 & 11 & 10 & a_{18} = \frac{31}{32} \\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \frac{19}{20} \\ [1ex] 5 & 19 & 1 & \text{N/A} & \text{N/A} & \text{N/A} \\ [1ex] \end{array}\] As $\left(j_1,j_2,j_3,j_4,j_5\right)=(1,2,8,18,30),$ the answer is $\sum_{i=1}^{5}{j_i}=\boxed{059}.$


Alternatively, from $(*)$ we can set \[\frac{m+18k'}{m+1+19k'}=\frac{t}{t+1}.\] We cross-multiply, rearrange, and apply Simon's Favorite Factoring Trick to get \[\left(k'+1\right)(t-18)=m-18.\] Since $k'+1\geq2,$ to find the smallest $k',$ we need $k'+1$ to be the smallest prime factor of $m-18.$ Now we continue with the last two paragraphs of the solution above.


Video Solution

https://youtu.be/oiUcYn1uYMM ~Math Problem Solving Skills

Video Solution by Punxsutawney Phil


See Also

2021 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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