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Mon 24 Nov, 2003 04:21 pm
You're standing blindfolded in front of a table. On the table are 128 coins with heads up and an X number of coins with the other side up. You're mission is to make 2 groups of coins with in each group the same number of coins with heads up. Remember you're blindfolded.
Good riddle! Will PM you the answer.
At the risk of being proved wrong, I would say that this one is impossible. Being blindfolded, you are limited to
1) counting
2) separating the coins into groups, or re-combining groups
3) turning one, some, or all coins over.
4) Combinations of 1) 2) and 3)
There seem to be no other possibilities than these.
None of these, as far as I can figure, can guarantee to result in two groups having the same number of heads showing for any value of 'X' ands for any random arrangement of the coins.
Can anyone suggest something I have missed? And does the fact that 128 is two to the seventh power have any relevance? (If not, why 128 and not some other number?)
It's actually very easy. The number 128 is crucial to the solution but you could reword the puzzle and use any number you liked.
If you were to turn over 128 coins randomly -- and keep that group separate from all the others on the table -- I think you would end up with two groups, both containing the same number of "heads up" coins.
Right?
Frank
It seems right to me. Isn't it always so simple once someone has told you how, eh? (I wish I was smart!)
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