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\begin{document}
\title{Iberoamerican 2021/3}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 87}
\maketitle
\section*{Problem}
Let $a_1$, $a_2$, $a_3$, \ldots
be a sequence of positive integers and let
$b_1$, $b_2$, $b_3$, \dots be the sequence of real numbers given by
\[ b_n = \dfrac{a_1a_2\cdots a_n}{a_1+a_2+\cdots + a_n},
\ \mbox{for}\ n\geq 1. \]
Show that, if there exists at least one term
among every million consecutive terms
of the sequence $b_1,b_2,b_3,\ldots$ that is an integer,
then there exists some $k$ such that $b_k > 2021^{2021}$.
\section*{Video}
\href{https://youtu.be/uePJQlZHhNs}{\texttt{https://youtu.be/Wx4-fscAMB8}}
\newpage
\section*{Solution}
Let $K = 10^{10^{100}}$ be a huge absolute constant.
We consider two cases.
The first case is captured in the following claim.
\begin{claim*}
Suppose that after the first $K$ terms,
at least one of every $K$ terms $a_i$ is greater than $1$.
Then $b_i$ is unbounded.
\end{claim*}
\begin{proof}
In that case, consider the first $n$ terms for some large $n$.
Let $M = \max(a_1, \dots, a_n)$.
We have that
\[ b_n \ge \frac{
2^{\left\lfloor 10^{-6} n - 1 \right\rfloor} M}
{M+M+\dots+M}
\ge \frac{1}{n}
2^{\left\lfloor 10^{-6} n - 1 \right\rfloor}
\]
since the product of every million terms is at least $2$
and one of the products is actually equal to $M$.
The right-hand side is unbounded in $n$, as desired.
\end{proof}
Thus, we may focus on the case where
$a_N = a_{N+1} = \dots = a_{N+K-1} = 1$ for some $N \ge K$.
Look at two indices $b_{N+i}$ and $b_{N+j}$ with $|j-i| \le 10^6$
and such that $b_i$ and $b_j$ are integers,
as promised by the problem condition.
We may write
\[ b_i = \frac{X}{Y+i}, \qquad b_j = \frac{X}{Y+j} \]
where $Y = a_1 + \dots + a_N \ge N \ge K$.
Then
\[ X \ge \lcm (Y+i, Y+j)
= \frac{(Y+i)(Y+j)}{\gcd(Y+i, Y+j)}
\ge \frac{1}{10^6} (Y+i)(Y+j) \]
which means that
\[ b_j \ge \frac{Y+j}{10^6} > \frac{K}{10^6} \]
as needed.
\end{document}