## Task $5 / 74$

Let $n$ be a natural number. How many decimal places (excluding trailing zeros) does the number $4^{-n}$ have?
The task is solved if it succeeds in writing the number $4^{-n}$ as a product of a natural number (which is not divisible by 10) and a power of ten with a negative exponent. Powers of ten always contain the prime factors 2 and 5 in equal numbers.

Thus, the following decomposition suggests itself:

$$
4^{-n}=2^{-2 n}=2^{-2 n} \cdot 5^{-2 n} \cdot 5^{2 n}=5^{2 n} \cdot 10^{-2 n}
$$

Since powers of 5 do not contain the prime factor 2, no power of 5 is divisible by 10 (each power of 5 ends with 5 as the last digit).

Consequently, the number $4^{-n}$ has exactly $2 n$ decimal places.