To find the number of whole numbers between 1 and 100 that use the digit "6" at least once, we can follow these steps:

1. Total numbers between 1 and 100: There are 100 numbers from 1 to 100 inclusive.

2. Numbers without the digit "6":

   • For a one-digit number, it can be any of 1-9 except 6, which leaves 8 options (1, 2, 3, 4, 5, 7, 8, 9).
   • For two-digit numbers, both digits must not be "6":
     • The first digit (tens place) can be 1-9 except 6, so 8 choices (1, 2, 3, 4, 5, 7, 8, 9).
     • The second digit (ones place) can be 0-9 except 6, so 9 choices (0, 1, 2, 3, 4, 5, 7, 8, 9).
   • Thus, the total count of two-digit numbers without "6" is $8 \times 9 = 72$.

3. Counting the number "100": The number "100" does not contain the digit "6".

4. Calculating numbers with at least one "6":

   • Numbers without "6" are 8 (one-digit) + 72 (two-digit) = 80.
   • So, the numbers containing at least one "6" = $100 - 80 = 20$.

Thus, there are 20 whole numbers between 1 and 100 that use the digit "6" at least once.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. How many whole numbers between 1 and 1000 use the digit "6" at least once?
2. How many whole numbers between 1 and 200 use the digit "7" at least once?
3. How would the count change if considering numbers without leading zeros?
4. How many whole numbers between 1 and 500 use at least one of the digits "3" or "6"?
5. What is the probability that a randomly selected number between 1 and 100 contains the digit "6"?

Tip: When dealing with digits in numbers, systematically consider each place value (units, tens, hundreds, etc.) for accurate counting!