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engg.manik
- Senior | Next Rank: 100 Posts
- Posts: 65
- Joined: Thu Jul 02, 2009 10:38 pm
- Location: Bangalore
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If I understand problem correctly, it seems a little strange....
There can't be just 3 boxes, because having a remainder of 5 would be foolish, because you could put 1 more in each box and still have remainder of 2, more optimal than 5.
How many boxes would prevent 5 balls from being distributed equally? 6 boxes.
So you had 77 balls, put 12 balls in each of 6 boxes, and had 5 left over.













