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\begin{document}
\title{Florida 2023}
\subtitle{Evan Chen}
\author{Twitch Solves ISL}
\date{Episode 113}
\maketitle

\section*{Problem}
Find all pairs of positive integers $(n,d)$ such that the following statement holds:
if $x$ is a positive integer for which $2^{x+n}$ and $5^x$ have the
same first decimal digit, that digit is equal to $d$.
(In particular, $d \le 9$.)

\section*{Video}
\href{https://www.youtube.com/watch?v=O2-1qI9SEHg&list=PLi6h8GM1FA6yHh4gDk_ZYezmncU1EJUmZ}{\texttt{https://youtu.be/O2-1qI9SEHg}}



\newpage

\section*{Solution}
Answer: $(2,6)$, $(4,1)$, and $(6,2)$.

First, note that \[ 2^{x+n} \cdot 5^x = 2^n \cdot 10^x. \]

To show these work, we start with the wonderful table
that shows what happens when
one multiplies two numbers with the same starting digit:
\begin{itemize}
  \ii $\ol{01{\dotso}} \le \ol{1{\dotso}} \cdot \ol{1{\dotso}} < \ol{04{\dotso}}$
  \ii $\ol{04{\dotso}} \le \ol{2{\dotso}} \cdot \ol{2{\dotso}} < \ol{09{\dotso}}$
  \ii $\ol{09{\dotso}} \le \ol{3{\dotso}} \cdot \ol{3{\dotso}} < \ol{16{\dotso}}$
  \ii $\ol{16{\dotso}} \le \ol{4{\dotso}} \cdot \ol{4{\dotso}} < \ol{25{\dotso}}$
  \ii $\ol{25{\dotso}} \le \ol{5{\dotso}} \cdot \ol{5{\dotso}} < \ol{36{\dotso}}$
  \ii $\ol{36{\dotso}} \le \ol{6{\dotso}} \cdot \ol{6{\dotso}} < \ol{49{\dotso}}$
  \ii $\ol{49{\dotso}} \le \ol{7{\dotso}} \cdot \ol{7{\dotso}} < \ol{64{\dotso}}$
  \ii $\ol{64{\dotso}} \le \ol{8{\dotso}} \cdot \ol{8{\dotso}} < \ol{81{\dotso}}$
  \ii $\ol{81{\dotso}} \le \ol{9{\dotso}} \cdot \ol{9{\dotso}} < \ol{100{\dotso}}$
\end{itemize}
Hence, if two numbers with the same leading digit
have product $4 \cdot 10^x$, for example, that leading digit is $2$ or $6$.
Moreover, the former case only occurs if the two numbers are both
exactly of the form $200\dotso00$.
So in the situation of the problem, if $2^{x+2} \cdot 5^x = 4 \cdot 10^x$
and $2^{x+2}$ and $5^x$ have the same leading digit,
that leading digit must be exactly $6$.
This shows that $(2,6)$ is a working pair.
The pairs $(4,1)$ and $(6,2)$ have exactly the same proof.

The converse is painful.
It relies on the following lemma:
\begin{lemma*}
  [Not JMO 2016/2]
  For any finite string of decimal digits,
  some power of $5$ begins with that string of digits.
\end{lemma*}
\begin{proof}
  The idea is to note that $\log_{10} 5$ is irrational,
  and hence for any nontrivial interval $I \subseteq [0,1]$
  there exists an integer $k$ such that
  the fractional part of $k \log_{10} 5 = \log_{10}(5^k)$ lies in $I$.
  (This is sometimes called Kronecker approximation theorem.)
\end{proof}

Now do some work: if $2^n$ is not a one- or two-digit perfect
square, there are two entries in the table earlier for which
the first two digits line up with the first two digits
of $2^n = \ol{d_1d_2\dots}$.
Then use Kronecker approximation to rig $5^x$ such that it starts
with some integer $s$ for which $s \approx \sqrt{d_1d_2.d_3d_4d_5\dots}$,
and then again with $s \approx \sqrt{d_1 . d_2d_3d_4 \dots}$.
(This is admittedly sketchy.)

\end{document}
