Counting Roman number so make addition instead of add

So is like this

First you have to know that had the unit (0 to 9), decena (10 to 99), and centena (100 to 999) and so on milhar, bilher, trillion…

And first you count the higher if had at least a number with centena start with the centena but if only ha the dezena or the unidade you start with the higher.

I + II + XXII = so count the dezena 2 times the X and after the unidades five of I that is V and the number is XXV

XLV + XCVII + LXI = counting is XC and with the X is C and you had already C (one hundred), and had the LX so until now you have 160 and countinuing you will count the unidades and is VV that is X more one number for the LXX and the others is ‘I’ tree times and you have III and the number is CLXXIII

You always went for the higher numbers and in sequence for the units the higher number is V and the other is I

For the dezenas is the L and the other is X

So another example

XL + XXI + LXXX + CXV + DIII =
1)The higher centena	D
2)and now still had one C	Until now you count DC
3)now counting the dezenas the higher is L	So until now the number is DCL
4)continuing counting the dezenas is	You had a XL that with a x you had the L that you cancel one 10 less with a 10 plus; now you continue counting 5 X that is the same that a L With the already the others ‘L’ you will had a LL=C and a L that is CL
5) counting the unidades is	You find a V and 4 ‘I’ So you count V(5) 6,7,8,9 that is nine or IV
And the number is:	DCCLIV

Another way of explain

XL + XXI + LXXX + CXV + DIII =

XL + XXI + LXXX + CXV + DIII =

D

XL + XXI + LXXX + CXV + DIII =

DC

XL + XXI + LXXX + CXV + DIII =

DCL

XL + XXI + LXXX + CXV + DIII =

DCLL

XL + XXI + LXXX + CXV + DIII =

DCLLL

XL + XXI + LXXX + CXV + DIII =

DCLLLIV that is DC(LL=C)LIV DCCLIV

Is like a calculator.

Rute Bezerra de Menezes Gondim