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#859578 - 12/31/12 03:50 PM Re: Roman Holiday [Re: Pandora]
Jema Offline
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Registered: 09/11/02
Posts: 13701
Loc: Virginia
7. II x XIV = XXVI + II (2 x 14 = 26 + 2)

II x IX = XVI + II (2 x 9 = 16 + 2)

newyear


Couldn't #1 just simply be II x II = IV?
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#859584 - 12/31/12 04:29 PM Re: Roman Holiday [Re: Pandora]
Pandora Offline

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Registered: 08/06/00
Posts: 48029
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.

smile12
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#859587 - 12/31/12 05:42 PM Re: Roman Holiday [Re: Pandora]
Jema Offline
Adept Boomer

Registered: 09/11/02
Posts: 13701
Loc: Virginia
OK. I misunderstood what "Remove one line segment from each side of the equal sign ..." meant. duh12

How about

7. II x XII = XXII + II (2 x 12 = 22 + 2)?
On each side of = the V becomes I.

HNY
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#859590 - 12/31/12 06:36 PM Re: Roman Holiday [Re: Pandora]
manxman Offline
Graduate Boomer

Registered: 07/23/02
Posts: 19416
Loc: Unionville
10. III x XXIX = XL + XLVII
II x XXIX = XI + XLVII newyear
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#859600 - 12/31/12 07:31 PM Re: Roman Holiday [Re: Pandora]
CanukDenis Online   content
Graduate Boomer

Registered: 11/26/00
Posts: 17380
Loc: Ottawa Ontario Canada
6. II x XXXI = XLV + XVII

I x XXXI = XIV + XVII : 1 * 31 = 14 + 17
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#859619 - 12/31/12 09:09 PM Re: Roman Holiday [Re: Pandora]
Pandora Offline

Sonic Boomer

Registered: 08/06/00
Posts: 48029
Loc: Upstate NY
Excellent solves all! newyear
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