The 3 digit numbers under 500 are 100 through 499.
1000 to 2999 inclusive so 2000 numbers.
Instead of going from 1 to 500, let's just subtract one from the set and come up with "How many 3's appear from 1 to 499?" Total possible combinations in this, including leading zeros, is 5*10*10, i.e. 5 possible in the first set(0,1,2,3,4), and 10 in the second two(0,1,2,,4,5,6,7,8,9). Instead of figuring out how many have a 3, lets figure out how many don't. Well there is 4 possibilities in the first (0,1,2,4), and nine in the second two (0,1,2,4,5,6,7,8,9). So, total = 5 * 10 * 10 = 500 Without 3 = 4 * 9 * 9 = 324 500-324 = 176 There are 176 numbers from 1 to 500 that have the digit 3 in them.
Answer : In total there are 22 numbers.
The digit 5 represents 5000 Each 1000 = 100 x 10 So 5000 = 500 x 10........there are thus 500 groups of 10.
The first digit can be any one of four (2, 4, 6, or 8). For each of those . . .The second digit can be any one of five (0, 2, 4, 6, or 8). For each of those . . .The third digit can be any one of five (0, 2, 4, 6, or 8). For each of those . . .The fourth digit can be any one of five (0, 2, 4, 6, or 8).Total number of possibilities = (4 x 5 x 5 x 5) = 500 .
If you include the number 3,000, there are exactly 2,001 4-digit numbers between 500 and 3000.
500 or so.
625 if numbers can have leading 0s, 500 otherwise.
400 of them.
There are 6,750 such numbers.
500
450... There are 500 odd numbers from 000 to 999 inclusive. From 000 to 099, there are 50 odd numbers. 500 - 50 = 450.
The three-digit numbers between 100 and 499 are what we need to consider. The smallest three-digit number is 100, and the largest is 499. To find the total, we calculate (499 - 100 + 1 = 400). Therefore, there are 400 three-digit numbers between 99 and 500.
200
200
9
500