There are nine numbers which contain only one digit. There are 90 numbers which contain two digits. There are 900 numbers which contain three digits. There is one number which contains four digits.
Therefore, the number of digits is equal to (9x1)+(90x2)+(900x3)+4 = 2893 digits. If this includes spaces, there would be 999 spaces, therefore there would be 3892 keystrokes.
21124
How many numbers are divisible by 9 between 5 and 1000²
There are 267 such numbers.
There are 90 such numbers.
There are 31 perfect square numbers between 1 and 1000 (including 1).
7,888,888,908
1*9+2*90+3*1=192
10 keystrokes are needed to type "Wichita, KS."
For only the digits: 5,888,896. More if there are commas as thousands separators (as in this answer), or spaces between numbers (999,999 of those).
21124
1 to a hundred - 13 if you do that 1-100 5 if you do that it varys
The prime numbers (factors) of 1000 are: 2 and 5
168 prime numbers under 1000.
How many numbers are divisible by 9 between 5 and 1000²
Both 10 and 1000 are individual whole numbers.
There are 90 palindromic numbers between 100 and 1000
There are 168 prime numbers between 1 & 1000.