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#859578
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12/31/12
03:50 PM
Re: Roman Holiday
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Re: Pandora
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Jema
Adept Boomer
Registered: 09/11/02
Posts: 11566
Loc: Virginia
7. II x XIV = XXVI + II (2 x 14 = 26 + 2)
II x IX = XVI + II (2 x 9 = 16 + 2)
Couldn't #1 just simply be II x II = IV?
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#859584
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12/31/12
04:29 PM
Re: Roman Holiday
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Re: Pandora
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Pandora
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Registered: 08/06/00
Posts: 43769
Loc: Upstate NY
Jema, you show two equations but you rearranged the numerals. This puzzle requires you to subtract one line from each side of the equation.
And yes, your #1 would work if you removed two lines from each side.
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#859587
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12/31/12
05:42 PM
Re: Roman Holiday
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Re: Pandora
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Jema
Adept Boomer
Registered: 09/11/02
Posts: 11566
Loc: Virginia
OK. I misunderstood what "Remove one line segment from each side of the equal sign ..." meant.
How about
7. II x XII = XXII + II (2 x 12 = 22 + 2)?
On each side of = the V becomes I.
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#859590
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12/31/12
06:36 PM
Re: Roman Holiday
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Re: Pandora
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manxman
Adept Boomer
Registered: 07/23/02
Posts: 10984
Loc: Markham, Ontario
10. III x XXIX = XL + XLVII
II x XXIX = XI + XLVII
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#859600
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12/31/12
07:31 PM
Re: Roman Holiday
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Re: Pandora
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CanukDenis
Adept Boomer
Registered: 11/26/00
Posts: 14164
Loc: Ottawa Ontario Canada
6. II x XXXI = XLV + XVII
I x XXXI = XIV + XVII : 1 * 31 = 14 + 17
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#859619
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12/31/12
09:09 PM
Re: Roman Holiday
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Re: Pandora
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Pandora
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Registered: 08/06/00
Posts: 43769
Loc: Upstate NY
Excellent solves all!
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