How many three-digit whole numbers have no 7's and no 9's as digits?
For the first digit, I cannot use 0, 7, or 9, so I have 7 choices: 1, 2, 3, 4, 5, 6, or 8. For the second digit, I cannot use 7 or 9, but I can use 0, so I have 8 choices: 0, 1, 2, 3, 4, 5, 6, or 8. For the third digit, I have the same 8 choices as the second digit. Therefore, the total number of three-digit numbers with no 7's and no 9's as digits is the product of the number of choices for each digit: 7 x 8 x 8. I can use the distributive property to simplify this calculation: 7 x 8 x 8 = 7 x (8 x 8) = 7 x 64 = 448. So there are 448 three-digit whole numbers that have no 7's and no 9's as digits. 