218. Find the last two digits of the power $7^{1995}$.
$\triangle$ Let's find the last two digits of the first few terms of the sequence ($7^{n}$): $07,49,43,01,07,49,43,01,07,49, \ldots$.

From this, we can see that these digits repeat when the term number increases by 4, generally by $4k (k \in N)$. Since

$$
1995=4 \cdot 498 + 3
$$

the last two digits of $7^{1995}$ coincide with the last two digits of $7^{3}$, and therefore form the number 43. Answer: 43.