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

Explanation

Let us consider 10 successive 7-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)" "9`,
where `a_(1),a_(2),a_(3),a_(4),a_(5) and a_(6)` are some digits, we see that half of these 10 numbers, i.e., 5 numbers have an even sum of digits.
the first digit `a_(1)` can assume 9 different values and each of the digits `a_(2),a_(3),a_(4),a_(5) and a_(6)` can assume 10 different values.
the last digit `a_(7)` can assume only 5 different values of which the sum of all digits is even.
`therefore` There are `9xx10^(5)xx5=45xx10^(5)`, 7-digit numbers the sum of whose digits is even.