Argentine National Olympiad 2015

Let $P=2\cdot 3\cdot 5\cdot \cdots \cdot 29$ be the product of the first 10 primes $2,3,5,7,11,13,17,19,23,29$. Observe that $P$ is the smallest unacceptable number. Apparently $P$ divides $100!$, and acceptable numbers are not divisible by $P$. Let $A$ remove $k_1$ stones on his first move. Because $100!$ is divisible by $P$ but $k_1$ is not, the number $n_1=100!-k_1$ of stones remaining is not divisible by $P$; in particular $n_1\ne 0$. So the remainder $r_1$ of $n_1\mathrm{mod}P$ satisfies $1\le r_1<P$. It follows that $r_1$ is acceptable as $P$ is the least unacceptable number. In addition $r_1$ is nonzero, so $B$ can make a legal move by taking $r_1$ stones. There remain $n_1-r_1$ stones, a quantity divisible by $P$.

Then, just like above, any move of $A$ yields a number $n_2$ of stones that is not divisible by $P$, and nonzero in particular. Its remainder $r_2\mathrm{mod}P$ is such that $1\le r_2<P$, so $B$ is able to remove $r_2$ stones and reach a position again where the number of stones is a multiple of $P$. Clearly $B$ can apply such moves at each step.

The number of stones decreases at each move, so the game ends with a win of one of the two players. $A$'s moves always leave a quantity not divisible by $P$, unlike $B$'s moves. Hence $A$ cannot take the last stone, meaning that $B$'s strategy guarantees him a win.