[Number Theory]
Problem: Let $X$ be the number of digits in the decimal expansion of $100^{1000^{10,000}}$, and let $Y$ be the number of digits in the decimal expansion of $1000^{10,000^{100,000}}$. Compute $\left\lfloor\log _{X} Y\right\rfloor$.

Solution: The number of digits of $n$ is $\lfloor\log n\rfloor+1$. Because $100^{1000^{10,000}}=\left(10^{2}\right)^{1000^{10,000}}, X=2$. $1000^{10,000}+1$. Similarly, $Y=3 \cdot 10,000^{100,000}+1$. Using the change-of-base formula,

$$
\begin{aligned}
\log _{X} Y=\frac{\log Y}{\log X} & \approx \frac{\log 3+\log 10,000^{100,000}}{\log 2+\log 1000^{10,000}} \\
& =\frac{\log 3+100,000 \log 10,000}{\log 2+10,000 \log 1000} \\
& =\frac{\log 3+100,000 \cdot 4}{\log 2+10,000 \cdot 3} \\
& =\frac{400,000+\log 3}{30,000+\log 2}
\end{aligned}
$$



Both $\log 3$ and $\log 2$ are tiny compared to the integers to which they are being added. If the quotient 400,000/30,000 were an integer (or extremely close to an integer), the values of these logarithms might matter, but $400,000 / 30,000=40 / 3=13 . \overline{3}$, so in this case, they are irrelevant. Hence

$$
\left\lfloor\log _{X} Y\right\rfloor=\left\lfloor\frac{400,000}{30,000}\right\rfloor=\left\lfloor\frac{40}{3}\right\rfloor=13
$$

Final answer: 13