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.

engg.manik wrote:Q -> Given that there are "x" boxes and boxes are filled with equal number of balls. If 3 boxes are filled with 12 balls each, then 5 balls are left after filling. Find the value of "x".

I agree that it's a very strange question, but we can answer it as (more or less) posted.

We know that 3 boxes are filled with 12 balls each and 5 balls are left over (maybe the boxes hold a maximum of 12 balls; nowhere does it say that we have to use all the balls in the 3 boxes).

So, we have 3*12 + 5 = 41 balls.

Now, assume that we must use all of the balls to fill our "x" boxes (although it doesn't say that explicitly, if it weren't the case it wouldn't be possible to answer the question). We know that every box has the same number of balls. Since 41 is a prime number, the only way we can fulfill the requirements is to put 1 ball in each of 41 boxes.

Therefore, x=41.