TITLE: Determine the number of functions $f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd.
QUESTION [5 upvotes]: Problems:Determine the number of functions $$f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$$ satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd. 
My Attempt: Each integer in domain has $4$ choices and therefore the total number of functions is $f$ is $4^{1999}.$ Since there are an equal number of functions that yield odd or even result, we can directly write that the number of functions satisfying the above condition to be $2*4^{1998}.$ I am unsure about this last claim and would like to prove it (if it is true.)Also is the answer to this problem correct?
Edit $1$: I tried to proceed further in the following manner: 
For a given mapping $f$ let the number of integers assigned to $2000,2001,2002$ amd $2003$ be $p,q,r$ and $s$ respectively, where $p,q,r,s\in \mathbb{Z}^+$. Therefore the sum in question can be written as $f(1)+f(2)+...+f(1999)=2000p+2001q+2002r+2003s.$ Clearly the sum will be odd iff $q+s$  is odd. Also note that $p+q+r+s=1999\Rightarrow p+r$ is even. On the other hand if we let $q+s$ as even then $p+r$ must be even. Now the number of values of $p,q,r$ and $s$ for which the above two lemmas hold is the same and therefore if we let $f$ to be the function which maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $q+s$ is even and $g$ to be the function  hich maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $p+q$ is even we will get a bijection.(Maybe!)

REPLY [3 votes]: Take an arbitrary image of $f(k)$ for $k\in\{1,\ldots,1998\}$ so you have for each $k$ a value from $2001$ to $2003$ and so you have $4^{1998}$ choices. Now for $f(1999)$ and since the sum would be odd, we have only two choices: so the desired number is $2\times 4^{1998}$