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Challenging Logic Puzzle

Challenging Logic Puzzle - 25 August

In front of you, there are 9 coins. They all look absolutely identical, but one of the coins is fake. However, you know that the fake coin is lighter than the rest, and in front of you is a balance scale. What is the least number of weightings you can use to find the counterfeit coin?

1. first take the coins in the multiple of 3 as soon as possible in this qn. 3 3 3 check first 3 3 if the coins are not equal divide another 3 coins into 1 1 1 check first 1 1 if the coins are not equal weight then the last one coin is the weightless one
common formula is 3^n where n is nearly power of 3.

1. 2. I think....

2. why 2

3. divide 3+3+3
weight 3-3 if equal odd one is in remaining 3
weight 1-1 if equal odd one is rem
So, Total 2 weights

4. Ya 2 is the minimum

5. The Answers is - 2.
Divide coin to 3 group-A,B,C. each 3 coin.

In First weight, A & B to find the lightest of A,B,C.

if weight of A is more than B. than B is lightest and vice-cersa.
if both are same weight than C is Lightest.

In Second weight, find the fault coin in lightest group. by wighting two coin same like above.

6. Minimum 2 weight

7. minimum 2 weight
maximum 6 weight

8. Minimum 2. Smaller-easier version of 12 coins, one is lighter/heavier, how do you find odd one out? (dont;'t know if it is lighter/heavier)

9. in only 3 weighings by the way!

10. 3.
First we take 4 coins on each pan of the balance leaving one outside. If both the pans are balanced, the left one is fake. If one pan is lighter than the other, the fake coin is among those four in it.So we weigh them dividing into two halves, i.e., 2 coins on each pan. One pan will be lighter, containing the fake coin. Then we weigh the third time: one coin on each pan. The lighter will be visible clearly.

11. its 3 only, We cant distribute into of 3 pair each because balancing exist only in 2 side

1. divide into 3 sets of coins...and balancing any 2 sets of coin...if it s equal then the 3rd set will contain the fake one...in 1st step we evaluate any one of 3 sets contain the fake one.....nxt step will be divide the 3 coins into 1,1,1....balance any 2 coins and rest the remainin one....if the balance s equal then the rest one will be the fake one.....

12. 2

ilovegrace.1210

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14. The answer is 1 weighting. If you happen to get the fake and weigh it against 1 real, you will see it's lighter. 1 weighting.

15. Actually because of the shit wording of the puzzle the answer is 1. Least number of weightings to find the couterfeit coin is 1 - you might get lucky and just weigh it next to one of the fine ones. If the question said to be CERTAIN of finding the counterfeit then the answer is two:

split nine into 3 groups of three (1) (2) and (3)

weigh (1) against (2)

2 possibilities:

possibility 1: (1)=(2)
therefore counterfeit is in group (3)
If any two coins from group (3) are weighed against each other there are another 2 possible outcomes:
(3(1))=(3(2)) therefore (3(3)) is counterfeit
(3(1))!=(3(2)) therefore whichever is lighter is the counterfeit

possibility 2: (1)!=(2)
therfore the lighter group contains counterfeit
follow same step as 1st possibility for the group containing the lighter coin to find the counterfeit

16. The answer is zero weightings. If one coin is indeed fake, it would be of a different make then, yes? And if it is of a different make, then it would produce a different sound when dropped, right? So I'd say, just drop each coin on a hard surface till you hear the one that sounds different.

17. Minimum is 1 if you are lucky
but 2 for a safe answer every time