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 2007-07-09, 18:11 #1 grandpascorpion     Jan 2005 Transdniestr 503 Posts Variation on a Martin Gardner puzzle You are given the task of finding counterfeit coins. Each batch of coins you must examine is already separated into "x" piles where x>1. "x" can be an arbitrarily large number. One of the "x" piles is contains only counterfeit coins. No other pile contains any counterfeits. Each pile may have a different number of coins but all must have "x" or more coins. Your only equipment is a modern one-panned scale. There is no practical limit as to how many coins you can put on the scale. Good coins weigh exactly 30 grams. Counterfeits weigh exactly 30.1 grams. The scale must be used to discern a weight difference. Also, the good and bad coins look exactly the same. ===================================================== How many weighings are necessary to determine the counterfeit pile? Please justify your answer. Last fiddled with by grandpascorpion on 2007-07-09 at 18:12
 2007-07-09, 18:47 #2 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts One weighing: one coin from pile1,2 from pile2...x from pile x. If the weight is W grams the conterfeit pile is 10(W-15x(x+1)) David
 2007-07-09, 18:57 #3 grandpascorpion     Jan 2005 Transdniestr 503 Posts Yep.
 2007-07-09, 20:12 #4 davieddy     "Lucan" Dec 2006 England 145128 Posts I guess that original puzzle was: 12 coins, one of which is either lighter or heavier than the rest. With a two pan balance, how many weighings to identify the counterfeit coin?
 2007-07-09, 20:23 #5 grandpascorpion     Jan 2005 Transdniestr 503 Posts This was based directly on his 10 piles of 10 coins problem but that's based on a simple puzzle like the one you describe.
 2007-07-09, 21:40 #6 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts Here's a variation: how many weighings on a single pan balance to resolve my 12 coin problem?
2007-07-12, 17:12   #7
mfgoode
Bronze Medalist

Jan 2004
Mumbai,India

22×33×19 Posts
Coins

Quote:
 Originally Posted by davieddy I guess that original puzzle was: 12 coins, one of which is either lighter or heavier than the rest. With a two pan balance, how many weighings to identify the counterfeit coin?

As per Martin Gardner the counterfeit coin can be identified and tell whether it is light or heavy in 3 weighings. He derives it by ternary numbers 0 , 1 , 2 and also by letters SILENT COWARD. I hope I have understood him correctly

Thanks for the reference grandpa and Davie.

Mally

2007-07-12, 18:15   #8
davieddy

"Lucan"
Dec 2006
England

2×3×13×83 Posts

Quote:
 Originally Posted by mfgoode As per Martin Gardner the counterfeit coin can be identified and tell whether it is light or heavy in 3 weighings. He derives it by ternary numbers 0 , 1 , 2 and also by letters SILENT COWARD. I hope I have understood him correctly Thanks for the reference grandpa and Davie. Mally
I've given this puzzle countless times to
pupils as a last day of term quiz, and yet I
still have to think hard about the solution in 3 weighings.
Each weighing requires precision and/or ingenuity.
I think a mnemonic is cheating. SILENT COWARD means
nothing to me.
The thing to remember is that we have 24 options
which must be resolved in 3 tests with 3 outcomes per test.

David

 2007-07-12, 18:37 #9 davieddy     "Lucan" Dec 2006 England 2×3×13×83 Posts BTW I've been trying to reply to your "Always an integer" post, and it quotes an earlier thread of yours instead. Not sure why. Anyway the answers to your queries are: a)x*integer is an integer because integer*integer=integer b)see books on "crooked E". David
2007-07-12, 19:51   #10
DJones

Oct 2006

1118 Posts

Quote:
 Originally Posted by davieddy Here's a variation: how many weighings on a single pan balance to resolve my 12 coin problem?
For a single pan scale, I'd say 5 weighings.
Three weighings of four coins each to determine whether the odd one out is heavier or lighter, which set of four it is in, and the weight of a normal coin.
One weighing of two coins to determine which pair the odd one out is in.
One weighing of one coin to determine the odd one out absolutely.

If the weight of a normal coin is already known, then I believe you only need four weighings;
One weighing of six coins, to determine which half the odd one out is in.
One weighing of three coins, to determine a set of three with the odd one out.
Two weighings (maximum) of one coin to find the odd one out.

I'm assuming a single-pan scale is an item on which you put objects and get a weight reading. I'm clarifying that as similar puzzles to this have used unusual terminology for scales and balances.

 2007-07-12, 20:25 #11 davieddy     "Lucan" Dec 2006 England 2·3·13·83 Posts I think you may be right. Goes to show that the two pan balance (though obsolete) had its advantages! David

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