20 digit number game, where the only the digits 1, 2, 3, 4, 5 are allowed. Albert and Charlie are taking turns writing digits of a 20-digit number, from left to right. Albert writes the first digit, Charlie writes the second, etc. until Charlie writes the last, 20th digit), using only digits 1, 2, 3, 4 or 5. Charlie wins if resulting number is divisible by 9 and Albert wins otherwise. Who wins if both players play optimally?
I started by attempting the game and saw that if Albert writes only 4 or 5 as digits, then Charlie wins because he can always make the sum of digits 9 by also writing a 5 or 4. I am not sure how to prove for cases where the number is smaller.
 A: Albert wins.  In what follows, all numbers and operations are modulo 9, since only divisibility by 9 matters.  Also, since we know that the sum of the digits determines the congruence class mod 9, we can forget about writing down the number and concentrate on the sum.
Suppose it's your turn, and you know that your that your opponent wants to to wants to make the sum a particular number at his next turn.  Let $d$ be the difference between his target and the current sum.  If $d=6,$ he cannot be stopped, since whatever number $x$ you say, he can say $d-6$.  However if $d$ is any other number, he can be blocked.  If $d$ is one of $7,8,0,1,$ you can say $1.$  If $d$ is one of $2,3,4,5,$ you can say $5$.
Now Charlie wants the sum to be $0$ at move $20,$ so he needs it to be $0-6=3$ at move $18,$ then $3-6=6$ at move $16,$ then $6-6=0$ at move $14,$ and so on.  Now we see that Charlie need the sum to be $3$ at the start of the game, so Albert wins by playing so as to deny Charlie any of his desired numbers.  In particular, Charlie wants the sum to be $0$ at move $2,$ so Albert can start by saying $1.$         
