By the pigeonhole principle, two of the numbers $1, 11, 111, \dots$ leave the same remainder when divided by $X$. Their difference, which is of the form $11\cdots100\cdots0$, is divisible by $X$. Therefore, we can transform $X$ to $11\cdots100\cdots0$, and then to $11\cdots1$ by deleting the zeros.

If the current number is $1$, we are done. Otherwise, we have
$$\underbrace{11\cdots1}_{k \text{ times}} \times 82 = \underbrace{911\cdots1}_{(k-2) \text{ times}} 02.$$
So we can transform $11\cdots1$ to $911\cdots102$, and then to $911\cdots12$. Next, note that
$$\underbrace{911\cdots1}_{m \text{ times}} 2 \times 9 = \underbrace{8200\cdots0}_{m \text{ times}} 8.$$
So we can transform the number to $8200\cdots08$, and then to $828$. Afterwards, we apply the operations as follows.
$$828 \xrightarrow{\times 25} 20700 \rightarrow 27 \xrightarrow{\times 4} 108 \rightarrow 18 \xrightarrow{\times 5} 90 \rightarrow 9$$
This completes the proof.