Open Contests

As $7\cdot 9\cdot 32=2016$, all turns are integral multiples of $\frac{1}{2016}$ of the full turn. Thus the carouse can be in at most 2016 distinct positions. It remains to show that all these positions are possible. For that, we show that the bears can turn the carouse by exactly $\frac{1}{2016}$ of the full turn. Then the same sequence of operations can be repeated to obtain also $\frac{2}{2016},\frac{3}{2016},\dots ,\frac{2016}{2016}$ of the full turn. Exactly $\frac{1}{2016}$ of the full turn is obtained, for instance, if bear dad turns the carouse once in one direction and both bear mum and bear cub turn the carouse once in the opposite direction since $\frac{1}{7}-\frac{1}{9}-\frac{1}{32}=\frac{288-224-63}{2016}=\frac{1}{2016}$.

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Alternative solution.

The carouse turns by an integral multiple of $\frac{1}{7}$ of the full turn due to bear dad, an integral multiple of $\frac{1}{9}$ of the full turn due to bear mum and an integral multiple of $\frac{1}{32}$ of the full turn due to bear cub. As the result, the carouse turns by $\frac{x}{7}+\frac{y}{9}+\frac{z}{32}$ of the full turn where $x,y,z$ are some integers. As $\frac{x}{7}+\frac{y}{9}+\frac{z}{32}=\frac{288x+224y+63z}{2016}$, the carouse can be turned only by integral multiples of $\frac{1}{2016}$ of the full turn. As $\text{gcd}(288,224,63)=\text{gcd}(9\cdot 32,7\cdot 32,7\cdot 9)=1$, there exist integral coefficients $a,b$ and $c$ such that $a\cdot 288+b\cdot 224+c\cdot 63=1$. Hence taking $x=na,y=nb$ and $z=nc$ for any integer $n$, the carouse turns by exactly $\frac{n}{2016}$ of the full turn. Hence all integral multiples of $\frac{1}{2016}$ of the full turn are possible, i.e., the carouse can be left in 2016 distinct positions.