How many of the natural numbers from 1 to 600, inclusive, contain the digit 5 at least once or the digit 6 at least once?

Guest Jun 11, 2021

#1**+3 **

Let's write each integer from 0 to 599 with 0-padding to 3 digits. So they are 000, 001, 002, ..., 599. Let $S$ be the set of integers from 0 to 599, inclusive, that contain neither the digit 5 nor 6. Then the answer we are seeking is $601-|S|$. (There are 601 numbers from 0 to 600 inclusive; the number 000 is in $S$; and the number 600 contains 6 at least once.)

Looking at it this way, numbers in $S$ can be constructed like this: Firstly, choose the leftmost digit from 0, 1, 2, 3, or 4. There are 5 choices. Secondly, choose the middle digit from 0, 1, 2, 3, 4, 7, 8, 9. There are 8 choices. Lastly, choose the rightmost digit from 0, 1, 2, 3, 4, 7, 8, 9. Again there are 8 choices. This gives $5\cdot 8\cdot 8= 320$ numbers in S.

So the answer to our question is $601-320=281$.

Bginner Jun 12, 2021