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 Replies: 467 / Views: 17,928 |
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Valued Member
Canada
287 Posts |
GRIVNA was used on the reverse in 1726 and 1727. (KM#171 )
All other years had GRIVENNIK on the reverse.
I learned something today. Good night all.
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Pillar of the Community
United States
602 Posts |
k, Tomorrow @ 7 then. WOLF
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Valued Member
Canada
287 Posts |
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Pillar of the Community
United States
4000 Posts |
Very cool.  |
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Valued Member
United States
439 Posts |
I have been ready through these contest and giveaways and have come to the unfortunate realization that I cannot possibly compete. I cannot answer one question. I will go back to the roll searching forums. Have fun..
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Pillar of the Community
 United States
2605 Posts |
 Question 6
There are 81 identically looking coins. We know that all but one are genuine, the counterfeit weighs less than a real coin. You have a balance scale (without marked weights) at your disposal. What is the minimum number of weighings that guarantees that we can find the counterfeit? I reiterate, you need to guarantee to find the fake with that number of weighings without hoping to get lucky.
Most important: explain how would you conduct those weighings to pinpoint the light coin. Illustration has been provided by an AH 1299 Zanzibar pysa. |
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Pillar of the Community
United States
602 Posts |
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Pillar of the Community
United States
2605 Posts |
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Pillar of the Community
United States
602 Posts |
Explaination; would put 2 on the scale and they should balance. So you have weighed two. If they are equal then replace one coin at a time until the balance is off. TA-DAAAAAA!
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Pillar of the Community
United States
1031 Posts |
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Valued Member
Canada
287 Posts |
Place 40 on either side of the scale.
If balanced, the counterfiet is on the table.
If not, take the lightest side and split it into 20/20
split again to 10/10
5/5
2/2
1/1
Total = 6
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Pillar of the Community
United States
2605 Posts |
Quote:
Most important: explain how would you conduct those weighings to pinpoint the light coin. Docsfishn, your answer is incomplete. |
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Pillar of the Community
United States
539 Posts |
assuming there is no chance of discovering the lighter coin until all have been weighed then my answer is 12.
you would have 4 weighings of 9 coins on each side, assuming none of those stacks tilts, then the last set of 9 contains the light coin. You would then take the stack of nine that contains the light coin, remove one and place on scale with a stack of 8 real coins. If you were lucky the stacks would be equal and the one you removed is the fake but you are not that lucky so... you remove 1 coin from each side for another additional 8 weighings (8-8, 7-7,6-6 etc) leaving one coin on each side with the lighter one being obvious.
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Pillar of the Community
United States
2605 Posts |
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Valued Member
Canada
287 Posts |
to be honest, I would double check my findings, so Doc is more correct.
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Replies: 467 / Views: 17,928 |
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Posted 02/03/2011 12:38 am
