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Too Many Balls


$2023$ balls are divided into several buckets such that no bucket contains more than $99$ balls. We can remove balls from any bucket or remove an entire bucket, as many times as we want. Prove that we can remove them in such a way that each of the remaining buckets will have an equal number of balls and the total number of remaining balls will be at least $100$.



Proof Based Problems  


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Solution

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For the sake of contradiction, assume that this is not possible.

So, there is not $k$ boxes with $\frac{100}{k}$ or more balls because taking these $k$ buckets keeping exactly $\lceil\frac{100}{k}\rceil$ balls satisfies the condition.

  •  So, there is no box with 100 or more balls.
  •  there is not 2 boxes with 50 or more balls.
  •  there is not 3 boxes with 34 or more balls.
  •  there is not 4 boxes with 25 or more balls.
  • there is not 5 boxes with 20 or more balls.
  • There is not 100 or more boxes with balls.


So, the highest amount of total balls is $99+49+33+24+19\cdot(99-4)=2010<2023$

Contradiction.

So, this is always possible.

This is a proof based problem added for learning purposes and does not accept submissions.

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