Well basically, you have 10,20,30,40,50,60,70,80,90,100=11 zeros and that is going to happen 8 times (210,220,230,...,780,790,800)
Then you have 101,102,103,...,109 for a total of 9 zeros which happens 7 times
So 8*11+7*9=151
Let F(n) be the number of occurrences of the digit zero in all the positive integers less than or equal to n. For example, F(9)=0, F(10)=1, and F(103)=14. With is F(800)?
Is there an easier way than to just count all of them to solve this problem? Please help!