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

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

#### The Romans did not have the Hindu-Arabic numbers. So we will solve out this operation exactly the way the Romans were calculating it, without using the Hindu-Arabic numbers.

### 1. Cancel the numerals having the same value but different signs.

#### The 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.

#### Cross out the common symbols in the numerals of different signs.

#### The 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 a group in additive notation - the largest symbol is to the left, down to the smallest to the right - to calculate the value simply add up the values of the symbols: XI = X + I = 10 + 1 = 11

#### IV is a group in subtractive notation - a smaller symbol precedes a larger one - to calculate the value of the group subtract the value of the first symbol from the value of 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. Replace the groups in subtractive notation.

#### Replace any groups in subtractive notation in the roman numerals; that is, "uncompact" them using only the additive notation.

#### A group in subtractive notation = a group of two numerals, one of a lower value preceding another of larger value, 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 of the group subtract the value of the first symbol from the value of the second.

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

#### CCCXIX:

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

#### CCCXIX = CCCXVIIII

### 4. Put the Roman numerals of the same sign together.

#### Catenate the positive numerals together.

#### CXX +

#### CCC +

#### CCCXVIIII =

#### CXXCCCCCCXVIIII

#### Catenate the negative numerals together.

#### V +

#### LV =

#### VLV

### 5. Sort the symbols in descending order of their values.

#### Sort the symbols in descending order of their values, from left-to-right, with the largest symbol to the left, down to the smallest to the right.

#### Sort out the symbols of the positive numerals:

#### CXXCCCCCCXVIIII =

#### CCCCCCCXXXVIIII

#### Sort out the symbols of the negative numerals:

#### VLV =

#### LVV

### 6. Cross out (remove) any symbols occurring in both strings.

#### CCCCCCCXXXVIIII -

#### LVV =

#### CCCCCCCXXXIIII -

#### LV

### 7. Again, cross out the common symbols in the numerals of different signs.

#### Using the additive notation, convert larger symbols to smaller ones and cross out the common symbols.

#### The symbols of larger value, converted to additive notation:

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

#### CCCCCCCXXXIIII -

#### LV =

#### CCCCCCLLXXVVIIII -

#### LV =

#### CCCCCCLXXVIIII

### 8. Combine the repeating symbols together, in groups.

#### Starting on the right end (smaller values), combine groups of the same symbols into larger ones.

#### CCCCCCLXXVIIII =

#### CCCCC CLXXVIIII =

#### DCLXXVIIII

### 9. Rewrite the repeating symbols.

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

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

#### DCLXXVIIII =

#### DCLXXIX

## The final answer: