SOLUTION: Find the sum of all even five-digit positive integers that are multiple of 5.

Since they are even, they are multiples of 2.  Since they
are also multiples of 5, they are multiples of 2×5 or 10.

So we want the sum of the multiples of 10 from 10000 to 99990,
inclusive,

10000 + 10010 + 10020 + ∙∙∙ + 99970 + 99980 + 99990

This sequence is the sequence of all 4 digit integers from 1000 to 9999:

1000, 1001, 1002, ..., 9997, 9998, 9999 

inclusive, with 0's annexed on the end of each.

So the number of 5-digit multiples of 10 is the same as the 
number of all 4-digit integers from 1000 to 9999.

There are 9999 4-digit integers from 1 through 9999, but 
we can't count the 999 integers from 1 through 999 which 
have fewer than 4 digits.  So since there are 9999-999 = 9000
4-digit numbers from 1000 to 9999, there are likewise 9000
multiples of 10 from 10000 to 99990, inclusive.

Using the sum formula with

a1 = 10000, n = 9000, an = a9000 = 99990.









Edwin