By counting numbers I am assuming you are talking about positive integers (whole numbers with no decimals). You are basically asking how many 5 digit integers there are that contain at least one 6. Let's find how many 5 digit numbers there are total and then subtract the amount that...

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By counting numbers I am assuming you are talking about positive integers (whole numbers with no decimals). You are basically asking how many 5 digit integers there are that contain at least one 6. Let's find how many 5 digit numbers there are total and then subtract the amount that do not contain a 6 to get our desired answer.

All 5 digit numbers range between 10000 and 99999. So subtracting the two numbers you get 99999-10000=90000 total integers containing 5 digits.

Now let's determine how many of these numbers do not contain at least one 6. The first four place holding digits can contain between a 0 and a 9 and the fifth place digit (the ten thousands) can contain between a 1 and a 9. If we eliminate the possibility of a 6 from all of these digits, the possible list of numbers for each digit drops by 1. We can multiply each of the digits together to get the desired total:

8*9*9*9*9=52488

This is the total number of 5 digit numbers that contain at least one 6. So subtract the two numbers from one another to get the total number of 5 digit numbers that contain at least one 6:

90000-52488=37512

**So the number of counting numbers that contain a six is 37512.**