An equivalent problem: For $ \mathbb{N} $ denoting the non-negative numbers, find the parity of the set $$ \left\{ (k,l,m)\in\mathbb{N}^3\setminus\{ (0,0,0)\}|2^k3^l5^m\le 2017\right\} $$

Let $A_1,A_2$ and $A_3$ denote the set of numbers less than or equal to $2017$ that are divisible by $2,3$ and $5$ respectively. Note that \begin{align*} \left| A_1\cup A_2\cup A_3\right| &=|A_1|+|A_2|+|A_3|-|A_1\cap A_2|-|A_2\cap A_3|-|A_3\cap A_1|+|A_1\cap A_2\cap A_3| \\ &=1008+672+403-336-134-201+67 \\ &=1479. \end{align*} Hence after $2017-1479=538$ turns, the player is destined to lose. Thus the person starting first loses assuming that both the players play optimally.

It was pointed out to me by one of the South African students that the question only asks if a winning strategy exists, and so, in principle, "The game clearly ends in a finite number of turns, and can not end in a draw, and both players have perfect information etc..., and so there is a winning strategy for either the first or the second player. Deka's winning strategy is to play as the player with a winning strategy and to follow that winning strategy." is a perfectly correct solution.

Lol, unless the jury feels sympathetic or humorously love his/her answer, I am sure the rubric asks to show "some" work for the full score aha

Let's use Euler's Totient function to find the number co-prime to 2,3,5 and less than 2017. The larget number that is a multiple of 2, 3, and 5 and less than 2017 is $\boxed{2010}$. Using this fact the prime factorization of 2010 is: $$2010 = 2\cdot 3 \cdot 5 \cdot 67$$ By using the previous preposition, we can get the number of the Co-prime numbers less than 2010 which is what we need to progress in the problem. $$\phi(2010) = 2010 \left(1-\frac{1}{2}\right) \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right) \left(1-\frac{1}{67}\right) = \boxed{528}$$ Then we know that for sure there are 528 numbers that are not multiples of 2, 3, and 5. However, we need to include the Co-prime number that are $$2010 < n < 2017$$and the numbers that are a multiple of 67 but not a multiple of 2, 3, and 5. These numbers are: $$S = \{67, 469, 737, 871, 1139, 1273, 1541, 1943, 2011, 2017 \}$$$$|S| = 10$$ Therefore, the final answer is: $$528 + 10 = \boxed{538}$$.

Mazen_hisham123 wrote: Let's use Euler's Totient function to find the number co-prime to 2,3,5 and less than 2017. The larget number that is a multiple of 2, 3, and 5 and less than 2017 is $\boxed{2010}$. Using this fact the prime factorization of 2010 is: $$2010 = 2\cdot 3 \cdot 5 \cdot 67$$ By using the previous preposition, we can get the number of the Co-prime numbers less than 2010 which is what we need to progress in the problem. $$\phi(n) = 2010 \left(1-\frac{1}{2}\right) \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right) \left(1-\frac{1}{67}\right) = \boxed{528}$$ Then we know that for sure there are 528 numbers that are not multiples of 2, 3, and 5. However, we need to include the Co-prime number that are $$2010 < n < 2017$$and the numbers that are a multiple of 67 but not a multiple of 2, 3, and 5. These numbers are: $$S = \{67, 469, 737, 871, 1139, 1273, 1541, 1943, 2011, 2017 \}$$$$|S| = 10$$ Therefore, the final answer is: $$528 + 10 = \boxed{538}$$. Brilliant move