7,8  [Pairings and groupings; bijections]

How many nine-digit numbers exist, the sum of whose digits is even

#
Let's break down nine-digit numbers into pairs of consecutive numbers: (100000000, 100000001), (100000000, 100000001), ... In each pair, the sum of the digits of the second number is 1 more than the sum of the digits of the first, so exactly one of them is even. Therefore, numbers with an even sum of digits make up exactly half of the total number of nine-digit numbers, which is $9 \cdot 10^{8}$ (see the solution to problem $\left.\underline{60336}\right)$.

## Answer

$45 \cdot 10^{7}$ numbers.