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)

 A: $$112359550561797752809 \cdot 99 = 11123595505617977528091\;.$$
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: 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.
A: 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 $\rm\ x = 10\:.$
