For which number does multiplying it by 99 add a 1 to each end of its decimal representation?

This was asked by my maths lecturer a couple of years ago and ive been wracking my brains ever since:

Find a number that, when multiplied by 99 will give the original number but with a 1 at the beginning and a 1 at the end.

For example: 42546254 * 99 would equal 1425462541
(it doesn't, but it illustrates what the answer would look like)

$$112359550561797752809 \cdot 99 = 11123595505617977528091\;.$$

If you put what you say into an equation, it reads

$$10^n+1+10x=99x\;,$$ $$10^n+1=89x\;.$$

Then you just need to find an $n$ such that $10^n\equiv 88\pmod{89}$, and $n=22$ works.

A down-to-earth way of finding the answer is to write the equation as $100x - x = 1x1$, or $1x1 + x = 100 x$. So if $x=abc\dots xyz$, then

 1abc...xyz1
+  abc...xyz
------------
=abc...xyz00

This means that the last digit $z$ must be 9, so we now have

          1
 1abc...xy91
+  abc...xy9
------------
=abc...xy900

And from this we see that y=0, so that

         11
 1abc...x091
+  abc...x09
------------
=abc...x0900

Hence x=8, and so on. After 21 steps you reach

 11123595505617977528091
+  112359550561797752809
------------------------
=11235955056179775280900

and then you're done.

It was shown that $\rm\ x = 112359550561797752809\:.$
Notice that $1/89\ =\ 0.0112359550561797752808988\ldots$

EXERCISE $\: $ Explain it (this, perhaps, is the point of the OP).

NOTE $\ $ This is closely connected with fibonacci numbers. Hint:

$\rm\quad\quad\quad x^n\ =\ f_n\ x + f_{n-1}\ (mod\ x^2-x-1)\ $

and note $\rm\ f_{11} = 89\ $ which is $\rm\ x^2-x-1\ $ for $\rm\ x = 10\:.$