You have 5565 balls. One of them is heavier than the others. What is the minimum number of weighs required to find this ball?

You have 5565 balls. One of them is heavier than the others. What is the minimum number of weighs required to find this ball?

it’s 62..!!!

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?

5565/3=1855

so there will be 3 grps of 1855,so 2 comparisons

then 1855/5=371

so 5 grp of 371 so 4 comparisons required

371/7=53

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..

65..