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- Apr 13, 2013

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At a clock (on which we have the positions $1,2, \dots, 12$) we place at position $1$ a blue ball and at position $2$ a red ball. At discrete times ($1,2,3, \dots$) we shift the two balls. Each time we shift the blue ball by three positions and the red ball by one position. Will the two balls ever meet at the same position?

I have thought the following.

Let $B$ be the position of the blue ball and $R$ the position of the red ball. Then $B=1+3t$ and $R=2+t$, for some $t \in \mathbb{N}$.

$B=R \Rightarrow 1+3t=2+t \Rightarrow 2t=1 \Rightarrow t=\frac{1}{2} \notin \mathbb{N}$.

Thus, the two balls will never meet at the same position. Am I right?