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Counterfeit Coins

rayyz's picture

Mr Smith has lots of pound coins, ten boxes in all. Each box contains 100 pound coins, but one box contains coins which are counterfeit and are slightly lighter, 1/16 of an ounce lighter to be exact. The problem lies in the fact that they all look identical, the only way to tell them apart is to weigh them. Mr Smith knows the correct weight for a box, but how many weighings are required to determine which box contains the counterfeit ones?

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jilfer's picture

jilfer said 3 ???? ...

Assumption: Mr Smith has a weighing balance
Step 1:
Split the boxes into 2sets with each set having 5 boxes and compare them by placing the set of boxes in each arm of the weighing balance. Since the defective box is lighter, take the 5 boxes which appear lighter in the balance.

Step2:
Again split the 5 boxes into two set with each set having 2 boxes. Compare the 2 sets in a weighing balance as before.
Result 1: If both are equal, the remaining 1 box is the effective one
Result 2:If they are not equal, then step 3.

Step 3: Keep the remaining 2 boxes in each arm of the weighing balance, the lighter one is the defective one

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gypsy gal's picture

gypsy gal said Jilfer.. ...

You re nto giving us a chance to think..lol Looks like you are coorect..

Ray is this answer correct?

 

rayyz's picture

rayyz said Jilfer & Gypsy Gal ...

Nopes.

It is possible in just one weighing!

Mr. Smith also has the liberty to use an electronic weighing machine. Now having 10 boxes of pound coins, he can't be that poor to use a weighing balance, could he!? LOL

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500 QL Points!!!
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jilfer's picture

jilfer said Gypsy gal ...

There is a similar puzzles (i think it was a 9 ball or maybe 12 balls)..which is way complicated....applied the same logic here...guess it correct...will have to wait for ray to confirm
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jilfer's picture

jilfer said gypsy gal...here is you ...

gypsy gal...here is you chance...my answer is wrong....so go for it
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gypsy gal's picture

gypsy gal said hmmm... ...

I dont know....nahi malum...did you get the answer Jilfer?

 

rayyz's picture

rayyz said The Solution ...

Like I said, it takes just one weighing.

Take one coin from the first box, two from the second and so on. Now when the coins are weighed, the number of 1/16ths light will tell us which box contains the counterfeits. For example if it was box 5, the weighing would be 5/16 too light.

Cheers!
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500 QL Points!!!
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jilfer's picture

jilfer said Rayz..you are giving the ...

Rayz..you are giving the answers to early...you should let us think for a couple of days, and come up with an answer...anyway good one....

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rayyz's picture

rayyz said Jilfer ...

Your line to Gypsy Gal made me think that you probably couldn't solve it.

Anyhow, my bad, I'll try hold on for few days before publishing the solutions.

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500 QL Points!!!
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jilfer's picture

jilfer said Rayz...I was just joking to ...

Rayz...I was just joking to gypsy...anyway it was a good one..i don't think i would have come up with that one..
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