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You have a pile of 24 coins. Twenty-three of these coins have the same weight, and one is heavier. Your task is to determine which coin is heavier and do so in the minimum number of weighings. You are given a beam balance (scale), which will compare the weight of any two sets of coins out of the total set of 24 coins. How many weighings are required to identify the heavier coin?
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suresh
[COLOR="Purple"]Correct me if iam wrong...
1.Divide the pile to 2(12 each)....Weigh ...Take the pile that weigh heavier
2.Divide that pile to two(6 each) ...and weigh again...Take the pile that weighs more
3.Divide that pile to two(3 each) ...and weigh again...Take the pile that weighs more
4.Compare the weight of 2 coins of the remaining three,,
If they r equal,the third one weighs more
If not ,the one that weighs more...Weighs more!!
So..i think we need 4 weighs...????[COLOR="Purple"][/COLOR][/COLOR]
Good Try Rose...Four attempts not a correct answer...
Try Again...I think u found easily....
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suresh
Divide the pile into 3. 8 coins each. weigh two sets of coin.
Take the one which weighs heavier. If both are of equal weight, then take the one that is not yet weighed.
Again divide that 8 coins into 3 sets. one having 2 coins and other having 3 coins.
weigh the 3 coin sets. Take the set that is heavier. If both are equal, take the 2 coin set.
Do the same again.
Thus [B]can be done in 3 weighings[/B].
Yes Barbie...Your answer is correct...
Good work and thanks for contributing...
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suresh