How many positive 3-digit integers contain only odd digits?

125

To determine the number of positive 3-digit integers that contain only odd digits, we need to consider the possible digits that can be used in each place value.

In the ones place, there are 5 odd digits to choose from: 1, 3, 5, 7, and 9. Any of these digits can be used.

In the tens place, there are also 5 odd digits to choose from: 1, 3, 5, 7, and 9. However, since we have already used one odd digit for the ones place, there are only 4 odd digits remaining.

In the hundreds place, we also have 4 odd digits to choose from since we have already used two odd digits in the previous places.

Therefore, the total number of positive 3-digit integers that contain only odd digits can be calculated as follows:

Number of possibilities = (Number of odd digits for ones place) x (Number of odd digits for tens place) x (Number of odd digits for hundreds place)
= 5 x 4 x 4
= 80

So, there are 80 positive 3-digit integers that contain only odd digits.

To find out how many positive 3-digit integers contain only odd digits, we need to consider the possible values for each digit.

Since we are looking for odd digits, we can consider the digits: 1, 3, 5, 7, and 9.

For the first digit, it cannot be zero since we need a positive integer. So we have 5 choices: 1, 3, 5, 7, or 9.

For the second digit, it can be any odd digit, including zero. So we again have 5 choices.

For the third digit, we also have 5 choices.

To find the total number of 3-digit integers containing only odd digits, we multiply the number of choices for each digit.

So the solution is: 5 choices for the first digit × 5 choices for the second digit × 5 choices for the third digit = 5 × 5 × 5 = 125.

Therefore, there are 125 positive 3-digit integers that contain only odd digits.

first digit has to be 1,3,5,7,9

as does second, and third.

number odd numbers= 5*5*5=125
and curiously, the number of even digits must be 4*5*5 (lead digit cannot be zero)=100
Now, containing one odd digit.
4*5*5+5*5*5+4*5*5=325 in two instances, first digit is only of four choices..
containg two odd digits
2*5*5*5 + 4*5*5 (second term is first digit is even)= 250+100

all of those should add to 9*10*10=900
checking: 350+325+100+125=900