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Topic: 1's worth a dollar ........ (Read 633 times) |
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ShadowLady
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1's worth a dollar ........
« on: Mar 6th, 2004, 1:23am » |
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G'day I'm from Australia and new to this site - so apologies if I have put this in a wrong site!! 
I have been given this riddle to solve but noone in so far has cracked the answer hence I am here asking the experts! 
here it is;
1's worth a dollar
100's worth three
10 is worth two bucks
What can they be?
Any suggestions? I need this riddle to complete a murder mystery game I am participating in!!
Thanks from Down Under
ShadowLady 
p.s i love your site!!
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Sir Col
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impudens simia et macrologus profundus fabulae
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Re: 1's worth a dollar ........
« Reply #1 on: Mar 6th, 2004, 3:09am » |
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There are a number of variations of this puzzle (contexts in which the expected answer lies), but they all come down to the same principle. Here are a few more clues that should help you solve it:
5 is worth $1.
31 is worth $2.
681 is worth $3.
Guess what 4206 is worth.
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mathschallenge.net / projecteuler.net
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Cathos
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Re: 1's worth a dollar ........
« Reply #2 on: Mar 6th, 2004, 12:27pm » |
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Interestingly, 5 is worth $1, but $5.00 is worth $5
(at least in the ones I've seen)
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Sum Arbor
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rmsgrey
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Re: 1's worth a dollar ........
« Reply #3 on: Mar 8th, 2004, 5:10am » |
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And that presumably leaves $2 as worth $2.
The version I met previously is obviously less sophisticated - $2 and $5.00 were both out of scope for evaluatable expressions.
I'm afraid I can't think of any really clever hints to prove I know what I'm talking about, so I'll just observe that ::Romans would have a much steeper price curve::
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Sir Col
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impudens simia et macrologus profundus fabulae
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Re: 1's worth a dollar ........
« Reply #4 on: Mar 8th, 2004, 6:18am » |
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Except with the Romans it would not be continuously increasing. To a Roman, 5 would cost less than 4 but as much as 1. Even more curiously, 1000 would cost much less than 999 (assuming you wrote it as Romans did, and not the way Europeans did) but as much as 1.
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rmsgrey
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Re: 1's worth a dollar ........
« Reply #5 on: Mar 8th, 2004, 6:56am » |
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True, but for arbitrarily large numbers, the Romans would go as O(n) while we just go as O(log n)
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