Starting from a positive integer $a$, we formed the numbers $b=2 a^{2}, c=2 b^{2}, d=2 c^{2}$. What could the number $a$ be if writing the decimal representations of the numbers $a, b, c$ in this order next to each other exactly gives the decimal representation of $d$?
According to the problem statement, $b=2 a^{2}, c=8 a^{4}, d=128 a^{8}$. We denote the number of digits of $b$ by $n$ and the number of digits of $c$ by $k$. Then we can write the following equation:

$$
a \cdot 10^{n+k}+2 a^{2} \cdot 10^{k}+8 a^{4}=128 a^{8}
$$

Dividing by $a \neq 0$ and rearranging:

$$
10^{n+k}=a\left(128 a^{6}-2 \cdot 10^{k}-8 a^{2}\right)
$$

The expression in the parentheses on the right side of the equation is an integer. This means that $a$ must be a divisor of $10^{n+k}$, so in the prime factorization of $a$, only 2 and 5 can appear.

We want to show that $a$ cannot be divisible by 5. According to the problem, the number $c \neq 0$ is the same as the last $k$ digits of the number $d$. This implies that $c$ and $d$ end with the same number of zeros. If $a$ were divisible by 5, then $c=8 a^{4}$ would be divisible by 10, meaning $c$ would end with a zero. On the other hand, since $d=2 c^{2}$, $d$ would have to end with at least twice as many zeros as $c$, which is impossible.

Thus, $a$ cannot be divisible by 5, so in its prime factorization, only 2 can appear. Let $a$ be some power of 2, say $t$, so $a=2^{t}$. Substituting this into (2) and rearranging:

$$
2^{n+k} 5^{n+k}=-2^{t+1+k} 5^{k}+2^{3 t+3}\left(2^{4 t+4}-1\right)
$$

We know that $c=8 a^{4}=2^{4 t+3}$ is a $k$-digit number, so

$$
2^{3 k-3}=8^{k-1} \leq 10^{k-1} \leq 2^{4 t+3}
$$

from which

$$
\begin{gathered}
3 k-3 \leq 4 t+3, \quad k \leq \frac{4 t+6}{3} \\
k+t+1 \leq \frac{4 t+6}{3}+t+1 \leq 2 t+2+t+1=3 t=3
\end{gathered}
$$

This inequality means that the right side of (3) is divisible by $2^{t+1+k}$. But then the left side must also be divisible by it, which is only possible if $n+k \geq k+t+1$, or if

$$
t \leq n-1
$$

Using the fact that $b=2 a^{2}=2^{2 t+1}$ is an $n$-digit number, from (5) we have

$$
2^{3 t} \leq 2^{3 n-3}=8^{n-1} \leq 10^{n-1} \leq 2^{2 t+1}
$$

from which

$$
3 t \leq 2 t+1
$$

so $t=0$ or $t=1$.

In the first case, $a=1, b=2, c=8, d=128$ is indeed a solution, while in the second case, $a=2, b=8, c=128$, $d=32768$ is not a solution. Therefore, the number $a$ could only be 1.