Variation on a Martin Gardner puzzle

[grandpascorpion]

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.

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How many weighings are necessary to determine the counterfeit pile? Please justify your answer.

[davieddy]

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

[grandpascorpion]

Yep.

[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?

[grandpascorpion]

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.

[davieddy]

Here's a variation:
how many weighings on a single pan balance to
resolve my 12 coin problem?

[mfgoode]

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

[davieddy]

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

[davieddy]

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

[DJones]

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.

[davieddy]

I think you may be right.
Goes to show that the two pan balance
(though obsolete) had its advantages!
David