How many positive integers in the set {50, 51,, . . . , 298, 299} do not contain any even digits?

If you do not mind please explain the steps along the way in detail.

Thank you

There are 299-50+1 = 250 numbers in the set.

Since half the numbers are even, that leaves 125 with odd last digit

There are now 5 numbers, beginning with 5...29, or 25 sets of 5 numbers. There are 13 sets with odd 2nd digit, and 12 sets with even 2nd digit. Throw those 60 away, leaving 65 starting with 5,7,9,...,29

So, now there are 15 2-digit numbers, and 50 3-digit numbers starting with 10,11,...,29

Half of those start with 2,4,...,28, so toss out those 25, leaving 15+25 = 40 numbers with all odd digits.

To determine how many positive integers in the set {50, 51, ..., 298, 299} do not contain any even digits, you can follow these steps:

Step 1: Identify the range of numbers
The given set is {50, 51, ..., 298, 299}. We need to consider the numbers from 50 to 299 (inclusive).

Step 2: Count the odd digits
We need to identify the odd digits, which are 1, 3, 5, 7, and 9. These are the only digits that we can use in our positive integers.

Step 3: Count the possibilities for each digit position
Start by analyzing each digit position in the numbers and counting the possibilities for each position. Since we are dealing with three-digit numbers, we will analyze each digit position from left to right.

For the hundreds digit:
The hundreds digit can take values from 1 to 2 (as the range is 50-299), so there are 2 possibilities for the hundreds digit.

For the tens digit:
The tens digit can take any odd digit (1, 3, 5, 7, or 9) since there are no restrictions. Thus, there are 5 possibilities for the tens digit.

For the units digit:
Similar to the tens digit, the units digit can also take any odd digit (1, 3, 5, 7, or 9). Again, there are 5 possibilities for the units digit.

Step 4: Multiply the number of possibilities
To determine the total number of positive integers in the given set that do not contain any even digits, we multiply the number of possibilities for each digit position.

Total possibilities = (number of possibilities for the hundreds digit) x (number of possibilities for the tens digit) x (number of possibilities for the units digit)
Total possibilities = 2 x 5 x 5

Step 5: Calculate the result
Now, we can calculate the result by multiplying the numbers together:

Total possibilities = 2 x 5 x 5 = 50

Therefore, there are a total of 50 positive integers in the set {50, 51, ..., 298, 299} that do not contain any even digits.