[Algebra]
Problem: Compute $\left\lfloor 100000(1.002)^{10}\right\rfloor$.

Solution: Consider the expansion of $(1.002)^{10}$ as $(1+0.002)^{10}$. Using the Binomial Theorem yields the following:

$$
(1+0.002)^{10}=1+\left(\begin{array}{c}
10 \\
1
\end{array}\right)(0.002)+\left(\begin{array}{c}
10 \\
2
\end{array}\right)(0.002)^{2}+\left(\begin{array}{c}
10 \\
3
\end{array}\right)(0.002)^{3}+\cdots+(0.002)^{10} .
$$

However, when $k>3$, the terms $\left(\begin{array}{c}10 \\ k\end{array}\right)(0.002)^{k}$ do not affect the final answer, because $0.002^{4}=$ $0.000000000016=\frac{16}{10^{12}}$, and the maximum binomial coefficient is $\left(\begin{array}{c}10 \\ 5\end{array}\right)=252$, so

$$
\left(\begin{array}{c}
10 \\
4
\end{array}\right)(0.002)^{4}+\left(\begin{array}{c}
10 \\
5
\end{array}\right)(0.002)^{5}+\cdots+(0.002)^{10}<\frac{252 \cdot 16}{10^{12}}+\frac{252 \cdot 16}{10^{12}}+\cdots+\frac{252 \cdot 16}{10^{12}},
$$

where the right side of the inequality contains seven terms, giving an upper bound of $\frac{7 \cdot 252 \cdot 16}{10^{12}}$. The numerator is approximately 28000 , but $\frac{28000}{10^{12}}=2.8 \times 10^{-8}$. So even when multiplied by $100000=10^{5}$, these terms contribute at most $3 \times 10^{-3}$ to the value of the expression before rounding.

The result of adding the first four terms $(k=0$ through $k=3)$ and multiplying by 100,000 is given by the following sum:

$$
100000+10(200)+45(0.4)+120(0.0008)=100000+2000+18+0.096=102018.096 .
$$

Then the desired quantity is $\lfloor 102018.096\rfloor=\mathbf{1 0 2 , 0 1 8}$.

Final answer: 102018