If you choose to include 0 as a whole number (debatable it is only theoretical and not existent) then 99 (01, 02, 03...) but if you include 00 then it is 100. If you wish to include negative numbers as these also are only two digits then it is 199 (assuming you included 00, if not then 198). But even after all of that and you didn't include 01, 02... and -01, -02... then it would be 180.
And finally, 0 is another answer because 0000036.00000000000000 is still a whole number but has an infinite number of zeros on the end and in front of the '36'. There is a complex answer to a simple question, enjoy!
525
The smallest such number is 100, the largest is 999, so the size of this set is:999 - 100 + 1 Another way to solve this: You have: * 9 options for the first digit * 10 options for the second digit * 10 options for the third digit Multiply all these numbers together.
Assuming that the first digit can't be zero: If repetition of digits is permitted: (5 x 6 x 6) = 180 numbers of 3 digits. If repetition of digits is not permitted: (5 x 5 x 4) = 100 numbers of 3 digits.
5*5*4*4 = 400
There are exactly 320 pages in 852 digits.
how many numbers exactly have 4 digits ? 8900, 8999, 9000, 9999
there are 89 twodigits in whole numbers
90
890
900 of them.
27
there are 899 whole numbers that have three digits.
89999 Any whole number above 99999 has 6 or more digits, and any whole number below 10000 has 4 or less digits. Since there are only whole numbers involved, it is a simple subtraction of 10000 from 99999.
Assuming that 001, 080, etc are not allowed (that is a leading zero or two is not permitted), the smallest number with exactly three digits is 100. The largest number with exactly three digits is 999. So there are 999 - 100 + 1 = 900 numbers with exactly three digits.
All the whole numbers between 99 and 1,000. Simple subtraction will give you a number, if that's what you meant to ask. 900, the numbers above are exclusive.
90 of them.
There are 90 of them.