If $X$ is larger than say ${10}^5$, then $B/X$ is at most ${10}^6$, meaning that there are at most one million multiples of $X$ to check. A "naive" solution that checks each and verifies that it is composed of the given digits is efficient enough.

If $X$ is at most ${10}^5$, then integers divided by $X$ give remainders no larger than ${10}^5-1$.

We will construct the function $f(n,prefix)$, the number of multiples of $X$ between $A$ and $B$, if we still have to place $n$ more digits (of $11$) and $prefix$ represents the digits already placed.

To calculate the values of this function, first observe that, for any $n$ and $prefix$, we can only form numbers between $prefix\cdot {10}^n$ and $prefix\cdot {10}^{n+1}-1$, inclusive. Based on this we recognise three cases:

• If all numbers are between $A$ and $B$, then the actual prefix doesn't matter when forming the rest of the number, just its remainder when divided by $X$. The limited range of the remainders reduces the state space and we can memoize the values of the function based on the remainder.
• If all numbers are less than $A$ or greater than $B$, then the value is zero.
• If some (but not all) numbers are between $A$ and $B$, then the value of the function doesn't depend only on the remainder of dividing the prefix by $X$, so it isn't possible to memoize the value of the function on the remainder. Luckily, there are few such cases (proportional to the length of the number).