TITLE: Pigeon Hole. 80 numbered balls
QUESTION [7 upvotes]: We have $80$ numbered balls(From $1$ to 80).Among which are $45$ blue and $35$ orange. Prove that at least two blue balls differ by $9$. For example $13$ and $22$ or $69$ and $78$.
So they can differ by $1,2,\ldots,79$. Hmm...

REPLY [3 votes]: Let us assume the contrary.  
Now, we can easily see that for every blue ball placed at $k$-th, we need to place an orange ball at the $(k+9)$-th place. So, we get $35$ pairs of blue-orange balls without violating our assumed statement and have occupied $70$ positions.  
But, from here here on, we are left with $10$ balls which are all blue balls. So, it is obvious by pigeon-hole that we will get atleast $2$ blue balls which differ by $9$ in whatever way we arrange the balls.