how many three-digit integers greater than 500 contain only odd digits
The first number has to be 5, 7, or 9
The second number has to be 1,3,5,7,or 9
and so on. The number of integers will be 3*5*?
To find the number of three-digit integers greater than 500 that contain only odd digits, we need to consider the possible values for each digit.
The first digit must be greater than 5, so it can be either 7 or 9 (since 5 is not odd).
The second and third digits can be any odd digits from 1 to 9.
Thus, there are 2 choices for the first digit (7 or 9), and 5 choices for each of the second and third digits (1, 3, 5, 7, or 9).
To find the total number of three-digit integers greater than 500 with only odd digits, we multiply the number of choices for each digit:
2 choices for the first digit × 5 choices for the second digit × 5 choices for the third digit = 2 × 5 × 5 = 50.
Therefore, there are 50 three-digit integers greater than 500 that contain only odd digits.
To find the number of three-digit integers greater than 500 that contain only odd digits, we can go through each digit position and count the number of possibilities.
For the hundreds place, since the integer must be greater than 500, the only possibility is 5.
For the tens place, any odd digit can be used, which are 1, 3, 5, 7, and 9. So, there are 5 possibilities.
For the units place, again, any odd digit can be used, which are 1, 3, 5, 7, and 9. So, there are 5 possibilities.
To find the total number of three-digit integers greater than 500 that contain only odd digits, we multiply the number of possibilities for each digit position:
Total = Number of possibilities for hundreds place x Number of possibilities for tens place x Number of possibilities for units place
= 1 (from above) x 5 x 5
= 25
Therefore, there are 25 three-digit integers greater than 500 that contain only odd digits.