(This is my very first question here! :-) ) So I encountered this fun problem yesterday and am feeling like sharing it with you. I almost solved it with the help of my computer but I guess there is a way to solve it without.
So the question is: find all the integer $a$ so that $a*99$ is equal to $1a1$. (The same number "$a$" between two 1)
This is what I've done so far (spoiler alert if you want to solve it all by yourself):
I quickly converted this equation into this one: $89*a -1 =10^n$ with $n$ equals to the number of numbers in $a$ plus one.
Then comes $a=(10^n +1)/89$
So now the problem is equivalent to finding the n so that $10^n +1$ can be divided by $89$. One can notice maybe that $89$ is a prime number. Fermat theorem gives us : $10^{88} -1 =0 [mod 89]$ but this doesn't seem to help :/
Conceptually speaking, I find difficult to convert those two ideas mathematically: -How searching integers solutions differs from searching any solutions. -How to use the fact that n is actually the number of numbers in $a$ plus one.
Hope you find the problem interesting and that some very bright solutions will come to your mind :-)
NB: using a computer, I have found some n that verify the equation but this is cheating somehow...
The observation that $10^{88}\equiv1\pmod{89}$ is very useful.
If $a$ is an $n$-digit number, the equation can be rewritten as $$ 99a=10^n+10a+1 $$ and so $$ a=\frac{10^n+1}{89} $$ so we need $10^n\equiv-1\pmod{89}$.
By Euler-Fermat, $10^{88}\equiv 1\pmod{89}$. Let's try and find the order of $10$ modulo $89$. We have $10^2\equiv11\pmod{89}$, $10^4\equiv32\pmod{89}$ and $10^{11}\equiv55\pmod{89}$, so $10^{44}\equiv1\pmod{89}$, but $10^{22}\equiv88\equiv-1\pmod{89}$. There's no need to look at $10^8$.
So the order is $44$ and we have also found a solution to $10^n\equiv-1\pmod{89}$, namely $n=22$.
Suppose $10^m\equiv-1\pmod{89}$; then $10^m\equiv10^{22}$, so $10^{m-22}\equiv1\pmod{89}$ and therefore $m\equiv22\pmod{44}$.
The solutions are the positive integers of the form $22+44k$.
The smallest solution is $$ \frac{10^{22}+1}{89}=112359550561797752809 $$ The next solution is $$ \frac{10^{66}+1}{89}= 11235955056179775280898876404494382022471910112359550561797752809 $$