For what natural number $n$ will the sum of the digits in the decimal representation of $1 / 5^{n}$ be 5?
Since $\frac{1}{5^{n}}=2^{n} \cdot \frac{1}{10^{n}}$, the sum of the digits of $\frac{1}{5^{n}}$ and $2^{n}$ are the same, because dividing by powers of 10 only changes the position of the decimal point and the number of 0 digits. Therefore, we will look for the natural exponent of 2 where the sum of the digits is 5.

By examining the cases $n=0,1,2,3,4,5$, we can see that for $2^{5}=32$, the sum of the digits is exactly 5. We claim that if $n>5$, then the sum of the digits of $2^{n}$ cannot be 5, meaning the only solution is $n=5$.

Suppose this is not true, and there is a natural number $n>5$ for which the sum of the digits of $2^{n}$ is 5. In this case, each digit is at most 4 (since $2^{n}$ has at least two digits), the number itself is even, and not divisible by 10, so the last digit is either 2 or 4.

If the last digit were 4, then the first digit would be 1, and all intermediate digits would be 0. Since $n>5$, $2^{n}$ is divisible by 8 - and even by 64 - so the number formed by the last three digits is also a multiple of 8. According to the above, this number could be 4, 14, or 104, of which only 104 is divisible by 8. Thus, 104 could be the 2-power in question, but 104 is not a power of 2.

If the last digit is 2, then - since $2^{n}$ is divisible by 4 - the number formed by the last two digits is also divisible by 4. Therefore, the second-to-last digit is 3 or 1. In the former case, the already known 32 results. If the second-to-last digit is 1, then it is preceded by an odd digit, because the number formed by the last three digits can only be divisible by 8 in this way. The sum of the digits is 5, so the last three digits are 112.

Proceeding similarly, from the fact that the number formed by the last four digits is divisible by 16, and the number formed by the last five digits is divisible by 32, we get that the fourth and fifth digits from the end must necessarily be 0 and 1, respectively. The sum of the digits so far is 5, but the resulting 10112 is not a power of 2.

We have thus shown that only for $n=5$ will the sum of the digits of $1 / 5^{n}$ in decimal form be 5.