Ok, i really need help with this riddle, here it goes...
A king has 8 gold coins, one of the coins is counterfeit and weighs less than the others. The king has only one balance scale. How can the king find the counterfeit coin using the least number of weighings?
please explain!
thanks so mucho!
2006-10-07 07:05:44 · 6 answers · asked by thumbsucker. 2 in Science & Mathematics ➔ Mathematics
Put 3 coins on each side of the scale (means 6 are on the scales, 2 are sat by the side).
If the scales balance: then you know one of the remaining 2 coins sat by the side must be counterfeit, so you put 1 coin on each side of the scale to find which one it the counterfeit one. That's two weighings to get your answer.
If the scales didn't balance: then you take any two of the coins which were on the lighter side of the scales and put 1 coin on each side of the scale (leave the third sat by the side). If they scales balance, the counterfeit is sat by the side. If the don't balance the counterfeit one is the lighter of the two. That's still only your second weighing.
Answer: Two weighings.
2006-10-07 07:19:17 · answer #1 · answered by nobby_piles 2 · 1⤊ 1⤋
Divide the 8 coins into 3 sets with 3, 3, and 2 in the sets. Put the 3 and 3 on the balance scale. If it balances, the counterfeit coin is in the set of two and another weighing of 1 vs. 1 will show the counterfeit. If the first weighing does not balance, take the lighter set of three and divide it into 3 sets, 1, 1, and 1. Weigh two of these against each other. If they balance the counterfeit is the other one. If they do not balance, the counterfeit is the lighter one. In any case, you only need two uses of the balance scale.
2006-10-07 07:14:34 · answer #2 · answered by schvan 2 · 0⤊ 1⤋
first put 4 coins on each side of the balance.
one side should weigh less.
remove the coins from the heavier side of the balance.
now divide your 4 remaining coins ( which were all on the lighter side of the balance)into 2 and 2
put 2 coins on one side and 2 coins on the other side (2nd time balance is being used)
again -- one side should weigh less than the other
take the two coins on that are on the lighter side
put one coin on one side and the other coin on the other side ( third time balance is being used)
the side of the balance which is lighter has the counterfiet coin
the balance was used three times!
hope this helps
2006-10-07 19:07:09 · answer #3 · answered by ilovemath_pi 2 · 0⤊ 0⤋
put 4 coins on one pan, 4 on the other. One pile will weigh less, so the coin must be among them.
Divide that pile into 2 pairs, and weigh them. One pair will weigh less, so weighing each coin in that pair will reveal the counterfeit coin.
So, 3 weighings are needed using this approach.
2006-10-07 07:10:56 · answer #4 · answered by James L 5 · 0⤊ 1⤋
positioned a million coin from stack a million on the size, 2 funds from stack 2 on the size... and so on. the burden should be more suitable than 55 grams. Subtract 55 grams from the burden the size shows. no matter if that's a million gram extra then the counterfeits are stack a million, no matter if that's 2 grams extra then the counterfeits are in stack 2, and so on.
2016-12-04 09:18:40 · answer #5 · answered by geiser 4 · 0⤊ 0⤋
he puts 4 in one side of the scale and 4 on the other side
then he need to pick the side with less wieght
2 on one side 2 on the other
then 1 with 1
2006-10-07 07:08:27 · answer #6 · answered by Anonymous · 0⤊ 0⤋