SOLUTION: All the odd numbers from 51 to 5001 are written. What is the total numbers of digit used?

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Question 911664: All the odd numbers from 51 to 5001 are written. What is the total numbers of digit used?
Found 2 solutions by AnlytcPhil, Edwin McCravy:
Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
See solution below.
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Note:
I redid this problem. For some strange reason I didn't notice that it said "odd 
numbers" and I did it for all the numbers.  I don't know why I didn't see it,
It says "odd" plain as day!  Anyway, my bad.  It's the same principle, only a
little harder.  Here's my approach:

2-digit numbers:

The sequence of odd numbers:

51, 53, 55, ..., 97, 99

have 2 digits each.  There are the same number of those as there are
numbers which are 1 less than them, which are these:

50, 52, 54, ..., 96, 98, 

There are the same number of those as there are
numbers which are half of them, which are these:

25, 26, 27, ..., 48, 49  

These are the 49 numbers 1 through 49, minus the 24 numbers 1 through 24

and 49-24 = 25

So there are 25 numbers with 2 digits each, so the number of digits in
the 2-digit numbers in the original sequence is 2x25 or 50.

---------------------------------------------
3-digit numbers:

The sequence of odd numbers:

101, 103, 105, ..., 997, 999

have 3 digits each.  There are the same number of those as there are
numbers which are 1 less than them, which are these:

100, 102, 104, ..., 996, 998, 

There are the same number of those as there are
numbers which are half of them, which are these:

50, 51, 52, ..., 498, 499  

These are the 499 numbers 1 through 499, minus the 49 numbers 1 through 49

and 499-49 = 450

So there are 450 numbers with 3 digits each, so the number of digits in
the 3-digit numbers in the original sequence is 3x450 or 1350.

---------------------------------------------
4-digit numbers:

The sequence of odd numbers:

1001, 1003, 1005, ..., 4999, 5001

have 4 digits each.  There are the same number of those as there are
numbers which are 1 less than them, which are these:

1000, 1002, 1004, ..., 4998, 5000, 

There are the same number of those as there are
numbers which are half of them, which are these:

500, 501, 502, ..., 2499, 2500  

These are the 2500 numbers 1 through 2500, minus the 499 numbers 1 through
499

and 2500-499 = 2001

So there are 2001 numbers with 4 digits each, so the number of digits in
the 4-digit numbers in the original sequence is 4x2001 or 8004.

---------------------------------------------
Grand total: 50 + 1350 + 8004 = 9404 digits.

Edwin


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