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Topic: Roman Numerals (Read 5986 times) |
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PSesulka
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Where there is a will there is a way
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Roman Numerals
« on: Dec 11th, 2002, 3:34pm » |
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I heard this one over the radio, its pretty interesting.
In the least amount of moves, make this Roman numeral equation be equal.
I - X = XI
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Garzahd
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Re: Roman Numerals
« Reply #1 on: Dec 11th, 2002, 4:01pm » |
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What's the definition of a "move"? Can we move a character, symbol, or what?
I've seen matchstick variants of this too; is this one?
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PSesulka
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Re: Roman Numerals
« Reply #2 on: Dec 11th, 2002, 4:55pm » |
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A move is exactly what you think it is. Changing something that is there, adding something that isn't there, moving a number from one spot to another.
I know the matchstick riddles you are talking about, it is not a variant.
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lukes new shoes
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Re: Roman Numerals
« Reply #3 on: Dec 11th, 2002, 10:46pm » |
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swap the 2 numerals on the left hand side of the equation with each other, do the same to the 2 numerals on the right side.
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PSesulka
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Re: Roman Numerals
« Reply #4 on: Dec 12th, 2002, 4:53am » |
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You can do it in less moves than that.
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SWF
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Re: Roman Numerals
« Reply #5 on: Dec 12th, 2002, 6:34am » |
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Zero moves if you look at it upside down:
IX = X - I
Or does that count as a move since the viewer must move?
For one move of a character, put the X on the right of = to the left:
XI - X = I
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James Fingas
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Re: Roman Numerals
« Reply #6 on: Dec 12th, 2002, 8:38am » |
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Maybe this isn't what you were hoping for:
Change '-' to '+'. Somehow I don't get that "wow, cool" feeling.
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Doc, I'm addicted to advice! What should I do?
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PSesulka
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Re: Roman Numerals
« Reply #7 on: Dec 12th, 2002, 10:54am » |
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James Fingas thats what most people say, but that would count as one move. And you are able to do it in less moves than one, SWF got the answer, least amount of moves as possible, and thats zero. The equation was already correct you just had to look at it in the correct way.
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