Maybe you mean the prime factor,or it's very very easy(just itself).

Yes, it should be "prime divisor" indeed.

Yes..... you're right... Now, it's corrected and thanks . Carlos Bravo Lima -PERU

It's not that difficult anyway. Since none of the digits can be even, $2$ cannot divide $B$. Also, since none of the digits can be $0$ or $5$, we know that $5$ cannot divide $B$. Now we proceed by contradiction. If the supposition is false, then $\exists n\in\mathbb{N}$ such that all of its digits belong to the set $\{1,3,7,9\}$ and $n=3^a7^b$ for some non-negative, integral $a$ and $b$. Now, note that $3^07^0=1$, and thus the ones digit is in $\{1,3,7,9\}$ and the tens digit is even. Now proceed by induction. If $n=100x+10y+z$ where $x\ge 0$, $y$ is even, and $z\in\{1,3,7,9\}$, then both $3n$ and $7n$ will contain an even tens-digit and either $1,3,7,$ or $9$ as a ones-digit. So by induction we have that for all $a,b\ge 0$ that $3^a7^b$ will contain an even-tens digit. But this violates the fact that none of the digits can be even; contradiction. Therefore the theorem must hold.

Arguing by contradiction, we have $n = 3^a 7^b$ for some $a, b$. Now, consider $n \pmod {20}.$ We know by casework that $$\{k, 3k, 7k, 9k \} \equiv \{1, 3, 7, 9\} \pmod {20}$$for $k = 3, 7$. Thus by induction $3^a 7^b \in \{1, 3, 7, 9\} \pmod {20}$, so $n$ has even tens digit, contradiction.