How many different seven digit numbers are there the sum of whose digits is even ?

Explanation

Let us consider 10 successive seven digit numbers
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)0`
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)1`
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)2`
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)3`
…………………….
`a_(1)a_(2)a_(3)a_(4)a_(5)a_(6)9`
Where , `a_(1),a_(2),a_(3) , ........ a_(6)` are some digits.
Here we observe that, half of these 10 number i.e., five numbers have an even sum of digits and five numbers have an odd sum of digits . The first digit `a_(1)`, can assume 9 different values, each of digits `a_(2),a_(3),a_(4),a_(5),a_(6)` can assume 10 different values, the last `a_(7)` , can assume only 5 different values for which the sum of all the digits is even.
Hence the required number of different seven digit numbers `=9xx10^(5)xx5=45xx10^(5)`.