Answer / monika

In 2 iterations we can find one coin that is heavier out of 8 coins.
First take any 2 coins out of 8 coins then remaining is 6 coins from this u can we can have 2 solution.
i) divide 6 coins into 2 parts of 3 each then weigh them, one set of 3 coins will be heavier if it has heavy coin then from that 3 coins take one out and weigh remaining two
if both are equal then the remaining coin is heavy.
ii) weigh 2 coins and which is heavy we can find out.

Answer / monika

Hi, Wareagle, if this is done then it takes 3 iterations, but my can be done in 2 iterations,
suppose if the heavy weight coin is in one of the 2 which i divided first then when u weigh 2 stacks of 3 coins in shows equal weight then we can leave that and weigh and the remaining 2 coins from this we can find which is heavier.

Answer / monika

In this there is no need to know that, we can find the coin by the solution i provided. If possible try to do it, take 8 coins and weigh them as i said u will get to know.

Answer / wareagle

Answer 1 is incorrect. Suppose the heavy coin is in the
first two coins you removed from the stack of 8.

1. Divide the coins into 2 groups of 4 and weigh each stack.
one stack will be heavier. 2.Take the heavier stack of 4
coins and divide into 2 stacks of 2 coins each. Weigh the 2
stacks. 3.One will be heavier. Take the 2 coins left and
weigh them. Take out the heavier of the 2 coins.

Answer / wareagle

But you can not be sure your method will work every time.
If I know where the heavy coin was I could do it with zero
iterations.

Answer / wareagle

But you can never be sure using your method. If I could be
sure where the heavy coin was I could do it with no
iterations.

wo kya cheez hai jo saal me 1 baar mahine me 2 baar hafte me 4 baar aur din me 6 baar aati hai

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