# Prove that This Property Does Not Hold for Any Other Pair of Digits.

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.

• @ThomasFinley As far as I can tell, you have correctly answered the first part. As for the second part, your interpretation of "Prove $n\nmid m$ when $m$ is made up of any other pair of digits" is somewhat inaccurate. It should instead be phrased something like "Prove there's at least one $n$ where $n\nmid m$ when $m$ is made up of any other pair of digits". As for how to prove this, I've already given a fairly detailed explanation in my earlier comment. Also, as for it being "sufficient", I believe it should be, but that would be up to whoever is reading your answer to determine. Jul 21, 2023 at 6:03
• @Prem, I'm looking at the book (exercise 50 on page 59), and that is the exact wording. (And, it's not ambiguous.) Jul 21, 2023 at 6:24
• Let's write $P_{a,b}(m)$ for the statement, $m$ contains only the digits $a$ and $b$. The problem is, For all $n$, for all $a$, there exists $m$ such that $P_{0,a}(m)$ and $n$ divides $m$. Show that, if $ab\ne0$, then it is not true that for all $n$ there exists $m$ such that $P_{a,b}(m)$ and $n$ divides $m$. To do the second part, all you need is a single value of $n$ for which there is no such $m$. And $n=10$ is such an $n$. Jul 21, 2023 at 6:29
• @Prem, "given $n$, there exists $m$" means "for every $n$, there exists $m$". Jul 21, 2023 at 6:41
• "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". "given $n$" means "for every $n$". @Prem Jul 21, 2023 at 6:48

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".