Question 9357: of 200 coins, 199 are the same weight and ine is lighter than the others. Given the same balance, explain how the lighter coin may be indentfied in no more than 5 weighings.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! In fact,thetotalnumber when limited to 5 weighings should be 243 instead
of 200. [Of course, bigger number takes more times of weighing. ]
The magic number for this kind of puzzle is 3. Instead of coins ,I prefer to use balls or eggs.
The key point is: for 3 balls, take two of them to weigh on the two sides of the balance , only once can determine the lighter one.
[This is the crucial and minimal requirement that you have to know.]
For 9 =3^2 balls, divide them into 3 groups. First time to weigh 3 and 3 balls on two sides of the balance
case 1.If they are equal then the lighter one is in the remaining three.
So, one more weigh we are done .
case 2.If two sides are unequal, the lighter ball must belong to the lighter side of the balance. And it also takes one more weighing.
Similarly, for 27 =3^3 balls, divide them into 3 groups. First time to weigh 9 and 9 balls on two sides of the balance
case 1.If they are equal then the lighter one is in the remaining nine. So, two more weighs we are done .
case 2.If the two sides are unequal, the lighter ball must belong to the lighter side. And it also takes two more weighs.
Hence, for 81 =3^4 balls, it needs 4 times of weighs by dividing them into 3 groups of 27 balls and compare two subgroups as the first weighing.
Finally, for 243 =3^5 balls, it needs 5 times of weighs by dividing them into 3 groups of 81 balls and compare two subgroups as the first weighing, then it takes 4 more times of weighing to find the lighter one in 81 balls.
This puzzle is not that hard, try to think about it or discuss with your friend.
If you are interested in similar puzzles, try to think if you don’t know the bad guy is lighter or heavier among 12 balls and the number of weighing is at most 3 times.
By the way,don't dig yourself into the silly question about how to
divide 200 balls at the first time.
Good luck.
Kenny
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