We write all the positive integers run together as follows: $123456789101112131415 . . .$ We write all the positive integers run together as follows:
$123456789101112131415 . . .$
What three digit number begins at the $2014th$ digit?
I was thinking number theory here. Modulus. 
Can someone give me hints, not answers?

As First $9$ numbers have single digit each and the next $99−9=90$ numbers have two digits each
$$1\cdot9+(99-9)\cdot2=189$$
Let the $x$ be the three digit number which makes $2013$ digits
$$\implies(x-99)\cdot3+189=2014-1$$
Clearly, $x+1$ will be the required three digit number
