Pertin A.

Find all numbers of the form $2^{n}$ ($n$ is natural) such that when the first digit of their decimal representation is erased, the result is again a power of two.
The ratio of two powers of two is less than 100. On the other hand, their difference has the form $2^{m}\left(2^{n}-1\right)$. Therefore, $n \leq 6$. This difference is divisible by 10, so $2^{n}-1$ is a multiple of 5. Hence, $n=4,2^{n}-1=15$. According to the condition, $15 \cdot 2^{m}=a \cdot 10^{k}$, where $a \leq 9$. Clearly, $k=1$, then $3 \cdot 2^{m-1} \leq 9$. Thus, $m=1$ or 2.

## Answer

32,64.