How many odd numbers between $0$ and $2018$ are there that the sum of its digits is an even number?

Question

The title is pretty much all the problem, I can't seem to figure it out, got told I might have to use factorials (I don't see how?)

Answer 1

Hint If $0 \leq x \leq 999$ then exactly one of the numbers $x$ or $x+1000$ has odd sum of the digits.

This tells you immediately how many such numbers are between $0$ and $1999$. After that there are only few more left to check.

Answer 2

If I understood correctly: $$\{1,3,5,7,9,11,\cdots 2017\}\\ \Rightarrow\{\{1\},\{3\},\{5\},\{7\},\{9\},\{1,1\},\{1,3\},\{1,5\},\{1,7\},\{1,9\}\cdots\{2,0,1,7\}\}$$ $$S=\{1,3,5,7,9,2,4,6,8,10,\cdots 10\}$$ Taking $S\mod 2$ gives: $$\{1,1,1,1,1,0,0,0,0,0,1,1,1,1,1\cdots1,1,1,1,1,0,0,0,0\}$$ The pattern is very clear, the $0$'s and $1$'s repeating every $5$ terms.

Since we have $\frac{2018}2$ odd numbers from $0$ to $2018$, we have $ (\frac{1009}2)=504.5\to504$ odd numbers with even digit sums.