SOLUTION: 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.

Algebra.Com
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

RELATED QUESTIONS

you are given 8 identical coins and a balance scale. seven of the coins weight the same,... (answered by mananth)
You are given eight golf balls. seven of the golf balls have the exact same weight but... (answered by macston)
given 12 pool balls each identical in shape and colour knowing that one of the pool balls (answered by jim_thompson5910)
Watermelon one is 2 kilograms lighter than watermelon two and 5 times lighter than... (answered by josgarithmetic)
Watermelon A is 2 kg lighter than watermelon B and it weighs one fifth of the weight of... (answered by josgarithmetic)
1) Given 12 coins such that exactly one of them is fake (lighter or heavier than the... (answered by Edwin McCravy)
Watermelon one is2 kg lighter than watermelon two and five times lighter than watermelon... (answered by Alan3354)
Two cartons weigh 3x-2 and 2x-3. If the average weight of the cartons is 10, the heavier... (answered by checkley79)
From a worksheet: Two objects weigh in at 2000kg. The heavier object is 240 less than 3 (answered by jojo14344)