Is there any winning strategy? 2015 and Game with marbles!!! Two players, Alex and Brad, take turns removing marbles from a jar which initially contains $2015$ marbles. Assume that on each turn the number of marbles withdrawn is a power of two.  If Alex has the first turn and the player who takes the last marble wins, is there a winning strategy for either of the players?

I try to solve it like the follow:
First) I assume that both player know the amount of marbles they took each other
Second) The problem say they pick power of 2 then the possible picks are $[1 , 2 , 4, 8, 16 ,32,64 128 , 256 , 512 , 1024]$ there are only $2015$ marbles.
Third) the game has 2 players.
Continue:
a)  The only odd pick is $1$ due to $2^0$ is $1$. The rest are odds
If player A pick odd (1) B must pick an even number and A continue pick even and also B until complete the $2015$.
And
If player A pick even (2, 4, etc) B must pick an odd (1) number and A continue pick even and also B until complete the $2015$.
I cannot see how can I use 2015 to set a winning rule, I also try use binary but I couldn’t do any advance.
Please, let me know what I am doing wrong and how I can continue. Thanks, Greg Martin and Shailesh for your help.
Note: I try to use brute fore (excel ;-) ) but I see that multiple of 3 are the wining but I can’t not put my assumption in words with the $2015$ number.
Thanks again.

I think this is essentially the same as what @PSPACE-Hard said, but here is my take on the winning strategy; 
A winning strategy revolves all around the number $3$. This is because;


*

*Every number is either a multiple of $3$, or is $1$ ($2^0$) or $2$ ($2^1$) below a multiple of $3$.

*Also, if either player is left with the number $3$, they are guaranteed to lose, because whether they choose $1$ or $2$, the other player is left with $1$ or $2$, and it is possible to win from either of these two numbers.


As a number that is a power of $2$ is just like any other number, it fits with point (1), so $2^n+1$ or $2^n+2$ are multiples of $3$. Obviously $2^n$ cannot be a power of $3$, given its prime factors will be just $2$.
As long as the player does this right up until $2^1+1$, they will leave the other player with $3$, which is a guaranteed victory given point (2).
Therefore, the strategy to win is to simply leave the other player with a number that fits $3n$, where $n$ is a whole number. The thing that makes this a guaranteed win, is that the other player will not be able to reply with a number that is divisible by $3$, because taking away a multiple of of $2$ (or the number $1$) from a multiple of $3$, you will be left with a number that isn't a multiple of $3$.
This means that the first player is guaranteed to win, as long as they;


*

*Play the number $2$ as their first number, leaving $2013$, which is divisible by $3$, and;

*Play tactically to ensure that after each of the other player's move, they play $1$ or $2$ to leave a number divisible by $3$.

