You have 5565 balls. One of them is heavier than the others. What is the minimum number of weighs required to find this ball?
A single weight which is equally massive as the heavy ball or the other balls. or we can simply compare the balls among themselves rather than a weight
A single weight which is equally massive as the heavy ball or the other balls.
3 times is enough.
is it 8?
so there will be 3 grps of 1855,so 2 comparisons
so 5 grp of 371 so 4 comparisons required
6 comparisons r required
53 a prime no can be further divided so 52 comparisons required total=2+4+6+52=64
How did you get 65? Try doing the question for 6 balls, 9 balls, and so on. You should find a pattern. The answer would the number of groups of 3 the total balls can be divided in.
plzz give us d ans..
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