171
90 of them.
450
there are no 3 digit tis two digit! :) * * * * * 90 of them.
a lot * * * * * Possibly, though that answer is relative. The correct answer is 900.
None. The only way for it to be possible would be 3 zeros which is not considered a 3 digit numbers.
450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.
This is only possible if one of the digits is equal to zero. There are 90 3-digit numbers with a zero in the 10's place, and 90 3-digit numbers with a zero in the 1's place - and 9 numbers that have both a zero in the 10's place and a zero in the 1's place; these would be counted double if you just add the first two. So, you get: 90 + 90 - 9 such numbers.
To find the number of three-digit even numbers where the leftmost digit cannot be zero, we consider the structure of a three-digit number represented as (ABC), where (A) is the leftmost digit, (B) is the middle digit, and (C) is the rightmost digit. The leftmost digit (A) can be any digit from 1 to 9 (9 options), the middle digit (B) can be any digit from 0 to 9 (10 options), and the rightmost digit (C) must be an even digit (0, 2, 4, 6, or 8, giving 5 options). Thus, the total number of three-digit even numbers is (9 \times 10 \times 5 = 450).
100 is a 3 digit number.
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
9,000,000 if there are no other requirements.
Nine million of them.