Roman Holiday

Posted by: Pandora

Roman Holiday - 12/30/12 06:56 PM

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

Posted by: CanukDenis

Re: Roman Holiday - 12/30/12 08:35 PM

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?
Posted by: Pandora

Re: Roman Holiday - 12/30/12 10:11 PM

It figures you'd think to do that. lol12 You cannot change the operations in this puzzle, just the operands.
Posted by: Flo NS

Re: Roman Holiday - 12/30/12 10:19 PM

5. VII x XI = LXXXV - VIII

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

Isn't this already true? I must not be understanding something.
Posted by: Pandora

Re: Roman Holiday - 12/30/12 10:36 PM

Yes, all of the equations start out true. You must remove a line from each side and still have a true (but different) equation.
Posted by: Sondi

Re: Roman Holiday - 12/30/12 11:06 PM

3. VII x IX = LXIII headscratch (I think!?)
Posted by: CCbomber

Re: Roman Holiday - 12/30/12 11:44 PM

8. II x III x XVII = LX + XLII

I x III x XVII = IX + XLII (=51) smile12
Posted by: CanukDenis

Re: Roman Holiday - 12/31/12 12:01 AM

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
Posted by: manxman

Re: Roman Holiday - 12/31/12 01:19 AM

9. IV x XIX = LXXXI - V
IV x XX = LXXXI - I grin12
Posted by: Lex

Re: Roman Holiday - 12/31/12 05:37 AM

2. LX = LXII - II kissy12 (I'm kissing ass and doing it right!)
Posted by: Flo NS

Re: Roman Holiday - 12/31/12 10:06 AM

thanks Pandora.

I'll stick with #5.

VII x XI = LXXXV - VIII

VI x XI = LXXIV - VIII

6x11=66 = 74-8 =66
Posted by: Pandora

Re: Roman Holiday - 12/31/12 10:08 AM

Excellent answers! Denis, I'll give you extra credit for creativity. And you're a good kisser Lex. thumbsup12
Posted by: CanukDenis

Re: Roman Holiday - 12/31/12 11:46 AM

Originally Posted By: Pandora
And you're a good kisser Lex. thumbsup12

YUK!! Stay away from me!!
Posted by: butterflybabe

Re: Roman Holiday - 12/31/12 12:14 PM

4: VI x IX = LI + III = 54
Posted by: Pandora

Re: Roman Holiday - 12/31/12 12:35 PM

Good solve, Annette! smile12
Posted by: Jema

Re: Roman Holiday - 12/31/12 03:50 PM

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?
Posted by: Pandora

Re: Roman Holiday - 12/31/12 04:29 PM

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
Posted by: Jema

Re: Roman Holiday - 12/31/12 05:42 PM

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
Posted by: manxman

Re: Roman Holiday - 12/31/12 06:36 PM

10. III x XXIX = XL + XLVII
II x XXIX = XI + XLVII newyear
Posted by: CanukDenis

Re: Roman Holiday - 12/31/12 07:31 PM

6. II x XXXI = XLV + XVII

I x XXXI = XIV + XVII : 1 * 31 = 14 + 17
Posted by: Pandora

Re: Roman Holiday - 12/31/12 09:09 PM

Excellent solves all! newyear