TITLE: How many integers from $43523$ to $93107$ contain at least one digit $7$
QUESTION [5 upvotes]: How many integers from $43523$ to $93107$ contain the digit $7$ at least once?
I know that if we had $43000$ to $93000$, we would subtract integers that do not contain digit $7$ from the total number:  $50000 - (5\times9\times 9\times 9 \times 9)$ 
But how to do it with the given numbers?

REPLY [7 votes]: First idea:  if you want the number of something from $42523$ through $93107$, it may be easier to count the ones from $0$ through $93107$ and subtract the ones from $0$ through $42522$. Let $N(n)$ be the number of numbers up to $n$ that contain at least one $7$.  You want $N(93107)-N(43522)$.    It helps to prefix the numbers with zero so all numbers have five digits. Second idea:  $N(93107)$ consists of all the numbers that start with $7$, which is $70000$ through $79999$, plus all the others that contain at least one $7$.  For the ones that start with $0$ through $8$, that is $8$ times as many as the four digit numbers that contain at least one $7$.  This suggests a recursive algorithm.  I leave the problem that you only want some of the numbers that start with $9$ to you.  The idea is the same.