Given a positive integer $n,$ prove that there is a positive integer $m$ that to base ten contains only the digits $0$ and $1$ such that $n|m.$ Prove that the same holds for digits $0$ and $2,$ or $0$ and $3,$ $\cdots,$ or $0$ and $9,$ but for no other pair of digits.
I tried solving it as follows:
We consider $n$ numbers $1,11,111,\cdots,\underbrace{ 111\cdots 1}_{\text{n 1's}}.$ We assume, neither of these numbers leave a remainder of $0$ when divided by $n.$ For if, there were such a number in the above list, we are done. So, the possible remainders on division by $n$ are $1,2,\cdots,n-1.$ So, we have, $n$ numbers and $n-1$ remainders. By PHP, there exists two numbers that leave the same remainder on dividing by $n.$ We call these two numbers in particular as $p=\frac{10^p-1}{9}$ and $q=\frac{10^q-1}{9}.$ Then, WLOG, let $p\gt q$ and we have, $m=\frac{10^p-10^q}{9}\equiv 0\pmod n$ and $m$ is in fact the required type of number.
Similarly, for a digit $2\leq i\leq 9$ we can proceed similarly by consider the list : $$i,ii,iii,\cdots, \underbrace {iii\cdots i}_{\text{n times}},$$ and then proceeding similarly as above.
But then, I don't know how to show, that this is not true "for no other pair of digits."
It seems they want us to show that, if $(a,b)$ is any other digit, $n$ won't divide it! This seems a little weird and difficult to show.
As far as I can tell, you've correctly answered the first part of the problem.
Regarding the second part, since it's asking about using a pair of decimal digits other than $(0,1), (0,2), \ldots, (0,9)$, this means it's a pair of decimal digits where neither of them is $0$. However, then with the first part of "Given a positive integer $n$", if we choose something like $n = 10$, all integral multiples of it will have, in base $10$, a $0$ as the final digit. Thus, there's no positive integer $m$ where $n\mid m$ and $m$ contains, in base $10$, only $2$ non-zero digits.
Regarding the question's phrasing, I don't consider it to be ambiguous because it requires that, regardless of which positive integer $n$ is "given", we can always find a corresponding $m$ meeting the stated conditions. Nonetheless, I suggest the first part could be made a bit clearer by, for example, using "Prove that, for any given positive integer $n$, there is a positive integer $m$ ..." instead, or even just by changing the "a" in "Given a positive ..." to "any".