There are three methods to do this.  Take your pick.

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Method 1:

First find the number of counting numbers greater than 1000 and lestt than 100000.

There are 99999 counting numbers from 1 to 99999, inclusive and we must subtract
the first 1000, leaving 99999-1000 = 98999.  Half that is 49499.5. Since the
first number is 1 and the last number is 99999, and they are both odd, then
there are 49499 even ones and 49500 odd ones.
Answer: 49500.


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Method 2;

4 digit numbers:
Choose the 1st digit 9 ways, {1,2,3,4,5,6,7,8,9}
Choose the 2nd digit 10 ways, {0,1,2,3,4,5,6,7,8,9} 
Choose the 3rd digit 10 ways, {0,1,2,3,4,5,6,7,8,9}
Choose the 4th digit 5 ways,  {1,3,5,7,9}

9x10x10x5 = 4500

5 digit numbers:
Choose the 1st digit 9 ways, {1,2,3,4,5,6,7,8,9}
Choose the 2nd digit 10 ways, {0,1,2,3,4,5,6,7,8,9} 
Choose the 3rd digit 10 ways, {0,1,2,3,4,5,6,7,8,9}
Choose the 4th digit 10 ways, {0,1,2,3,4,5,6,7,8,9}
Choose the 1st digit 5 ways,  {1,3,5,7,9}

9x10x10x10x5 = 45000

Total:  4500+45000 = 49500

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Method 3:

The odd numbers from 1001 through 49999 form an arithmetic
sequence with , , and d=2.  So we use
the formula:

 




 

Edwin