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.
