# $9999\ldots9\cdot9999\ldots9$ will always contain exactly one $8$?

Will a number consisting of only the digit $9$, multiplied with another number consisting of only the digit $9$, always result in a number that contains exactly one $8$ digit, and how can one know that this will always be true?

Well, the multiplication can be written as $$(10^x-1)(10^y-1)=10^{(x+y)}-(10^x+10^y)+1$$ so the result will be $$1000\ldots000-(1000\ldots00+1000\ldots000)+1=1000\ldots000-100\ldots001000\ldots000+1=9999\ldots999899\ldots99999\ldots0000+1=9999\ldots999899\ldots99999\ldots0001$$
Such a number is of the form $(10^m-1)(10^n-1)=10^{(m+n)}-(10^m+10^n)+1, \text{such that}\ m,n>0.$ It is clear that only the $max{(m,n)}^{th}$ digit from right will be 8.
• $999\cdot 99=98901$. Did you mean min? – John Habert Mar 10 '14 at 15:24
• @John Habert: I think he means from the right and starting from $0$. – user2345215 Mar 10 '14 at 15:27
• @user2345215 Interestingly, it works both ways as max from right starting with $0$ and as min from left. But which is more natural? For example, if we do $99999\cdot 9 = 899991$, then do you really want to refer to the $8$ as the $5^{\mathrm{th}}$ digit? – John Habert Mar 10 '14 at 15:36
• @John Habert: I was just trying to defend the $\max$, but yeah, $m+n-\max=\min$ and it's more natural do count it your way. – user2345215 Mar 10 '14 at 15:39