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#859440 - 12/30/12 06:56 PM Roman Holiday
Pandora Offline

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Registered: 08/06/00
Posts: 45782
Loc: Upstate NY
Some members of the Roman numerals below want to take a holiday, but they don't want to cause any of the equations to become untrue. Remove one line segment from each side of the equal sign so that the resulting equation remains true. (When you remove one line from a V or an X, you may rotate the remaining line slightly.)

-1. II x VII = XIV
-2. LIX = LXII - III
-3. VIII x IX = LXXII
-4. VII x IX = LI + XII
-5. VII x XI = LXXXV - VIII
-6. II x XXXI = XLV + XVII
-7. II x XIV = XXVI + II
-8. II x III x XVII = LX + XLII
-9. IV x XIX = LXXXI - V
-10. III x XXIX = XL + XLVII

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#859452 - 12/30/12 08:35 PM Re: Roman Holiday [Re: Pandora]
CanukDenis Offline
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Registered: 11/26/00
Posts: 15267
Loc: Ottawa Ontario Canada
1. II x VII = XIV
II x -II = -IV

Removed line segment from the V and X,
and arranged remaining line segment into a minus sign:
is that ok?
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#859459 - 12/30/12 10:11 PM Re: Roman Holiday [Re: Pandora]
Pandora Offline

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It figures you'd think to do that. lol12 You cannot change the operations in this puzzle, just the operands.
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#859460 - 12/30/12 10:19 PM Re: Roman Holiday [Re: Pandora]
Flo NS Offline
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Registered: 04/11/06
Posts: 4362
Loc: Nova Scotia
5. VII x XI = LXXXV - VIII

7x11(77) = 85-8(77)

Isn't this already true? I must not be understanding something.


Edited by Flo NS (12/30/12 10:22 PM)

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#859461 - 12/30/12 10:36 PM Re: Roman Holiday [Re: Pandora]
Pandora Offline

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Registered: 08/06/00
Posts: 45782
Loc: Upstate NY
Yes, all of the equations start out true. You must remove a line from each side and still have a true (but different) equation.
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#859464 - 12/30/12 11:06 PM Re: Roman Holiday [Re: Pandora]
Sondi Online   content
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Registered: 08/15/05
Posts: 3605
Loc: California
3. VII x IX = LXIII headscratch (I think!?)

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#859467 - 12/30/12 11:44 PM Re: Roman Holiday [Re: Pandora]
CCbomber Offline
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Registered: 01/16/03
Posts: 3277
Loc: Mojave desert, California
8. II x III x XVII = LX + XLII

I x III x XVII = IX + XLII (=51) smile12

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#859472 - 12/31/12 12:01 AM Re: Roman Holiday [Re: Pandora]
CanukDenis Offline
Graduate Boomer

Registered: 11/26/00
Posts: 15267
Loc: Ottawa Ontario Canada
Originally Posted By: Pandora
You cannot change the operations in this puzzle, just the operands.

But I didn't change the operations: the "x" is still there;
changing VII to -II is changing the operand from 7 to -2


1. II x VII = XIV
II x VI = XII ; the V on right becomes a I

I think that's the answer you expect.
But I prefer my original yes
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#859477 - 12/31/12 01:19 AM Re: Roman Holiday [Re: Pandora]
manxman Offline
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Registered: 07/23/02
Posts: 15340
Loc: Unionville
9. IV x XIX = LXXXI - V
IV x XX = LXXXI - I grin12
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#859491 - 12/31/12 05:37 AM Re: Roman Holiday [Re: Pandora]
Lex Offline
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Registered: 03/07/04
Posts: 5108
Loc: Isle of Man
2. LX = LXII - II kissy12 (I'm kissing ass and doing it right!)
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#859509 - 12/31/12 10:06 AM Re: Roman Holiday [Re: Pandora]
Flo NS Offline
Addicted Boomer

Registered: 04/11/06
Posts: 4362
Loc: Nova Scotia
thanks Pandora.

I'll stick with #5.

VII x XI = LXXXV - VIII

VI x XI = LXXIV - VIII

6x11=66 = 74-8 =66

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#859510 - 12/31/12 10:08 AM Re: Roman Holiday [Re: Pandora]
Pandora Offline

Sonic Boomer

Registered: 08/06/00
Posts: 45782
Loc: Upstate NY
Excellent answers! Denis, I'll give you extra credit for creativity. And you're a good kisser Lex. thumbsup12
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#859532 - 12/31/12 11:46 AM Re: Roman Holiday [Re: Pandora]
CanukDenis Offline
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Registered: 11/26/00
Posts: 15267
Loc: Ottawa Ontario Canada
Originally Posted By: Pandora
And you're a good kisser Lex. thumbsup12

YUK!! Stay away from me!!
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#859541 - 12/31/12 12:14 PM Re: Roman Holiday [Re: Pandora]
butterflybabe Offline
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Registered: 07/27/05
Posts: 2170
Loc: Mt Prospect , IL
4: VI x IX = LI + III = 54
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#859547 - 12/31/12 12:35 PM Re: Roman Holiday [Re: Pandora]
Pandora Offline

Sonic Boomer

Registered: 08/06/00
Posts: 45782
Loc: Upstate NY
Good solve, Annette! smile12
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