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lite
11-26-2006, 11:41 PM
hey all
i remembered in both the exams there was a question on the romanian no.
like :
VII means :
a. 7
b. 2
c. ...
d. ...


i know it seems easy but one of the numbers were hard like CXII
i tried to look it up but i couldn`t get it .. please any resources ? any web sites ?
thanks in advance

fafy
11-27-2006, 12:02 AM
I=1
V=5
X=10
L=50
C=100
D=500
M=1000

Add each number if their value decrease from left to right. Eg: MCXVI=1000+100+10+5+1=1116
Subtract their value if their value increase. Eg: CIX= 100+10-1=109

fizzbizz
11-27-2006, 12:05 AM
SymbolValue
I (http://en.wikipedia.org/wiki/I) or i 1 or one (http://en.wikipedia.org/wiki/1_%28number%29) (unus)
V (http://en.wikipedia.org/wiki/V) or v 5 or five (http://en.wikipedia.org/wiki/5_%28number%29) (quinque)
X (http://en.wikipedia.org/wiki/X) or x is10 or ten (http://en.wikipedia.org/wiki/10_%28number%29) (decem)
L (http://en.wikipedia.org/wiki/L) or l is 50 or fifty (http://en.wikipedia.org/wiki/50_%28number%29)
C (http://en.wikipedia.org/wiki/C) or c is 100 or one hundred (http://en.wikipedia.org/wiki/100_%28number%29) (centum)
D (http://en.wikipedia.org/wiki/D) or d is 500 or five hundred (http://en.wikipedia.org/wiki/500_%28number%29)
M (http://en.wikipedia.org/wiki/M) or m is 1000 or one thousand (http://en.wikipedia.org/wiki/1000_%28number%29) (mille)

For large numbers (five thousand (http://en.wikipedia.org/wiki/5000_%28number%29) and above), a bar is placed above a base numeral to indicate multiplication (http://en.wikipedia.org/wiki/Multiplication) by 1000:

an underline X (X) is ten million

Rules regarding Roman numerals often state that a symbol representing 10 to the power x may not precede any symbol larger than 10to the power x+1. For example, C cannot be preceded by I or V, only by X (or, of course, by a symbol representing a value equal to or larger than C). Thus, one should represent the number "ninety-nine" as XCIX, not as the "shortcut" IC. However, these rules are not universally followed.

XLⅩ40 (http://en.wikipedia.org/wiki/40_%28number%29), LⅬ50 (http://en.wikipedia.org/wiki/50_%28number%29), LX60 (http://en.wikipedia.org/wiki/60_%28number%29), LXXⅩ70 (http://en.wikipedia.org/wiki/70_%28number%29) LXXX80 (http://en.wikipedia.org/wiki/80_%28number%29), XCⅩⅭ90 (http://en.wikipedia.org/wiki/90_%28number%29), XCIXⅩⅭⅠⅩ99 (http://en.wikipedia.org/wiki/99_%28number%29), CⅭ100 (http://en.wikipedia.org/wiki/100_%28number%29), CCⅭⅭ200 (http://en.wikipedia.org/wiki/200_%28number%29), CDⅭⅮ400 (http://en.wikipedia.org/wiki/400_%28number%29), DⅮ500 (http://en.wikipedia.org/wiki/500_%28number%29), DCLXVIⅠ666 (http://en.wikipedia.org/wiki/666_%28number%29) ,CMⅭⅯ900 (http://en.wikipedia.org/wiki/900_%28number%29), MⅯ1000 (http://en.wikipedia.org/wiki/1000_%28number%29), MIX=1009, MCMXLV1945 (http://en.wikipedia.org/wiki/1000_%28number%29), MCMXCIX1999 (http://en.wikipedia.org/wiki/1000_%28number%29)

fizzbizz
11-27-2006, 12:13 AM
sheesh what did i paste :crazy:
fafy's is simpler:tup:

lite
11-27-2006, 12:25 AM
thanks very much fafy & fizzbizz :tup:

khadija
11-27-2006, 05:44 AM
I=1
V=5
X=10
L=50
C=100
D=500
M=1000

Add each number if their value decrease from left to right. Eg: MCXVI=1000+100+10+5+1=1116
Subtract their value if their value increase. Eg: CIX= 100+10-1=109
FAFY,

could you please explain how to calculate these.
MCDXCII and MCMXCV it is respectively 1492 and 1995
and this one please CMXCIX

khadija
11-27-2006, 06:33 AM
:hmm:

fizzbizz
11-27-2006, 08:03 AM
FAFY,

could you please explain how to calculate these.
MCDXCII and MCMXCV it is respectively 1492 and 1995
and this one please CMXCIX
lemme try
1000+CD=500-100+XC=100-10 +1+1=1492
1000+ CM= 1000-100+XC=100-10+5=1995
CM=1000-100 + XC 100-10+ IX 10-1 =999

mtvua
11-27-2006, 12:34 PM
http://www.factmonster.com/ipka/A0769547.html

fafy
11-27-2006, 05:30 PM
Fizzbizz is correct and the site is very good.

khadija
11-27-2006, 05:46 PM
lemme try
1000+CD=500-100+XC=100-10 +1+1=1492
1000+ CM= 1000-100+XC=100-10+5=1995
CM=1000-100 + XC 100-10+ IX 10-1 =999

good job lady,[clap] [clap] :grad: thanks alot