#### Let's learn by an example:

## CCCXXXVIII + XC + CCCXLIII + CCCXIX - CXVII - CXLVIII - LVI - XC = ?

#### 338 + 90 + 343 + 319 - 117 - 148 - 56 - 90 = ? the Romans did not have the Arabic numbers

## CCCXXXVIII + XC + CCCXLIII + CCCXIX - CXVII - CXLVIII - LVI - XC = ?

### 1. Cancel equal but different signs numerals.

#### Matching numerals / symbols are colored the same.

#### CCCXXXVIII + XC + CCCXLIII + CCCXIX - CXVII - CXLVIII - LVI - XC =

#### CCCXXXVIII + CCCXLIII + CCCXIX - CXVII - CXLVIII - LVI

### 2. Simplify the operation right away by crossing out common symbols in the numerals of different signs.

#### Matching numerals / symbols are colored the same.

#### CCCXXXVIII + CCCXLIII + CCCXIX - CXVII - CXLVIII - LVI =

#### CXX + CCC + CCCXIX - V - LV

#### For example, the numerals XI and IV:

#### XI is an additive group - largest symbol on the left, down to the smallest on the right - to calculate the value add up the symbols: XI = X + I = 10 + 1 = 11;

#### IV is a subtractive group - a smaller symbol precedes a larger one - to calculate the value subtract the first symbol from the second: IV = V - I = 5 - 1 = 4.

#### When subtracting the two numerals, XI - IV = (uncompact the subtractive) XI - IIII = X - III = VIIIII - III = VII (= 7), CORRECT. But if you cross out the common symbol regardless of the fact that it is part of a subtractive group: XI - IV = X - V = VV - V = V (= 5), WRONG.

### 3. Substitute for any subtractives in the roman numerals; that is, "uncompact" them using only the additive notation.

#### Subtractive notation = a group of two numerals, one of a lower value preceding another larger one, the only allowed ones are: IV, IX, XL, XC, CD, CM, M(V), M(X), (X)(L), (X)(C), (C)(D), (C)(M) - to calculate the value subtract the first symbol from the second.

#### Additive notation = a group of two or more numerals, equal or sorted in descending order from high to low, largest symbol on the left, down to the smallest on the right - to calculate the value add up the symbols.

#### CCCXIX:

#### IX = X - I = V IIIII - I = V IIII = VIIII

#### CCCXIX = CCCXVIIII

### 4. Put the Roman numerals together, catenate them.

#### Catenate all the positive numerals together.

#### CXX +

#### CCC +

#### CCCXVIIII =

#### CXXCCCCCCXVIIII

#### Catenate all the negative numerals together.

#### V +

#### LV =

#### VLV

### 5. Sort the symbols in order from left-to-right with the largest symbol on the left, down to the smallest on the right.

#### Sort the symbols of the positive numerals.

#### CXXCCCCCCXVIIII =

#### CCCCCCCXXXVIIII

#### Sort the symbols of the negative numerals.

#### VLV =

#### LVV

### 6. Simply "cross out" (remove) any symbols occurring in both values.

#### CCCCCCCXXXVIIII -

#### LVV =

#### CCCCCCCXXXIIII -

#### LV

### 7. Using only the additive notation, convert larger value symbols to some smaller ones, a necessary step in order to cross out all the common symbols.

#### Replacements used, in additive notation:

#### C = LL; X = VV;

#### CCCCCCCXXXIIII -

#### LV =

#### CCCCCCLLXXVVIIII -

#### LV =

#### CCCCCCLXXVIIII

### 8. Starting with the right end (smaller values), combine groups of the same symbols that can make a "larger" one and substitute the single larger one.

#### CCCCCCLXXVIIII =

#### CCCCC CLXXVIIII =

#### DCLXXVIIII

### 9. Rewrite the symbols written in excessive additive notation by using the substractive notation.

#### The numeral I should not repeat itself more than 3 times in a row, rewrite:

#### DCLXXVIIII =

#### DCLXXIX

## Final answer: