A Totally Crappy System for Doing Math

I’ve actually been fascinated by Roman numerals for a long time. You know them: Columbus discovered America in MCDXCII, and my birth year is MCMLX. Even Google knows Roman numerals these days—you can just type a number followed by “in roman numerals” and it gives you the Roman version.

What’s so fascinating about the whole system is how incredibly difficult it is. Hard to read, yes—but especially hard to do math with. A simple question like: 

If it’s the year MMXXIV (2024), and I was born in MCMLX (1960), how old am I? The formula is: MMXXIV – MCMLX = ?, and that’s surprisingly hard to solve.

I use the year of 2024 here, because it explains the process a little better. 2025 - 1960 is a simpler equation ;-)

And then there are sums like CCCXII + DCCCLXI, or XXI × XVII. Translate these into our modern Arabic numerals and the math becomes simple. We all learned how to handle 312 + 861 and 21 × 17 in primary school. It turns out that these sums are far easier to solve in our Arabic numeral system than in the Roman one. Of course, the Romans couldn’t just convert the number into Arabic digits, do the math, and convert it back again.

What’s especially intriguing is that the Romans built monumental architecture, levied taxes, constructed toll roads, and conducted trade—all using a totally crappy system for doing math.

You can find some info online (also an innovation the Romans didn’t have!) about Roman-style arithmetic. It turns out that addition and subtraction were fairly straightforward and even kind of elegant. Then comes a surprising fact: the Romans could multiply accurately, but there’s no indication they understood why their method worked. Division was a nightmare. Only the most trivial problems—dividing by 2, 3, or 4—seem to have had any solutions. For more complex divisions like LXIII ÷ VII (63 ÷ 7 = 9), there seems to be no reliable method.

Further down, I explain some Roman techniques for doing math. You don’t have to read that unless you’re really curious. Like I said before: a totally crappy system for doing math, but much more elegant than I initially thought. An entire empire was built on it. But it was a lot of work. The use of Arabic numerals and Arabic algebra was a true innovation that made all of this much easier—with huge consequences...

Lets do the math: Simple addition first

The Roman system is based on the order of the symbols you use. If a smaller number comes before a larger one, you have to subtract it; if the larger number comes first, you add them together. So

MMXXIV is 1000 + 1000 + 10 + 10 + 5 - 1 = 2024

MCMLX is 1000 - 100 + 1000 + 50 + 10 = 1960.

The Romans used a compressed form, like how we write it. Let’s call this the Normal Form. In addition, you can also use various expanded forms to represent the same number. So MCMLX can also be written as

MDCCCCLX (=1000+500+100+100+100+100+50+10=1960),

but also as

MCMXXXXXX (=1000-100+1000+10+10+10+10+10+10=1960),

and many other forms. It is customary to write the shortest form, which is why the usual (Normal Form) is MCMLX.

Once it's clear that you can switch between the Normal Form and various expanded forms, adding and subtracting becomes easy. The trick is to first convert the number to an expanded form in which no subtractions occur—so no smaller number before a larger one. Let’s call this the Simple Form. Then you just count up all the symbols and convert it back to the Normal Form.

So:

CCCXII + DCCCLXI (312 + 861 in the Normal Form)

becomes, in Simple Form (in this case it’s already simple):

CCCXII + DCCCLXI (312 + 861 in the Simple Form).

Add all the symbols together:

CCCXII + DCCCLXI  = DCCCCCCLXXIII,

and convert that to the Normal Form: Six C’s can be replaced by DC, giving DDCLXXIII, and two D’s can be replaced by M, giving

MCLXXIII (1000+100+50+10+10+1+1+1 = 1173).

Once this is clear, subtraction also becomes easier

First convert both numbers to Simple Form, then convert the first number into a form that has more of each symbol than the second number. Then cross out the number of symbols in the second number from the first, and convert the result back to Normal Form.

MMXXIV – MCMLX (2024 – 1960)

becomes:

MMXXIIII – MDCCCCLX (both converted to Simple Form),

then:

MDCCCCCXXIIII – MDCCCCLX (one M from the first number is replaced with DCCCCC),

then:

MDCCCCLLXXIIII – MDCCCCLX (one C is replaced with LL in the first number).

Now the first number has more of every symbol than the second.

Now we can cross out: the M in both numbers, the D in both, and so on. What’s left is LI from the first number, which we can’t cancel.

MDCCCCLLXXIIII – MDCCCCLX = LXIII = 50 + 10 + 4 = 64,

and that’s indeed the outcome of 2024 – 1960 = 64.

Okay, for everyone who’s still reading: halving numbers.

The Romans didn’t have fractions, so halving an odd number would leave a remainder. Once the Roman subtraction system is clear, halving becomes easy too: First convert to Simple Form, then expand it until every symbol appears an even number of times. Then remove half of each, and convert back to Normal Form.

Example:

Get the halve of CLXXVII (177). 

First, the Simple Form—this already is.

From left to right: There's only one C, so that's not even. Replace C with LL: 

Get the halve of LLLXXVII.

Cross out one L, move one L to the answer. Left with LXXVII. One L again—not even.

Get the halve of LLLXXVII: 

Answer: L, remainder: LXXVII

In the remainder, convert L to XXXXX: total now is XXXXXXXXVII.

Get the halve of LLLXXVII: 

Answer: L, remainder: XXXXXXXVII

Now, group X’s in pairs: move one to the answer, one to discard.

There are 7 X’s, so 3 go to the answer, 3 are discarded, 1 left over.

Get the halve of LLLXXVII: 

Answer: LXXX, remainder: XVII

Answer now contains LXXX. Leftover: XVII.

In the remainder we can see:

• One X: not even. Convert to VVVII.

• Cross out one V, move one to the answer. Left: VII.

• One V: not even. Convert V to IIIII. Now we have 7 I’s, and the answer so far is LXXXV.

• Now pair up the I’s: 3 go to the answer, 3 are discarded, 1 left.

Get the halve of LLLXXVII: 

Answer: LXXXVIII, remainder: I 

Final answer: LXXXVIII (= 88) with a remainder of 1.

Another correct outcome.

So, now we can already solve quite a few calculations with the Roman system. We can add, subtract, halve, and double. It’s a bit laborious, but it works. Clearly, careful calculators were held in high regard among the Romans.

Multiplications, another round of challenges

Now the fascinating part: the Romans discovered that you can do any multiplication using a table of doubling and halving. It took us much longer to understand why their method actually works—we now explain it using binary systems, which we only really mastered in the 19th century. 

The technique is as follows (to keep it readable, I’ll use Arabic numerals):

21 × 17; that’s XXI × XVII.

We build a table with two columns. The column on the left starts with the first term (21), the column on the right starts with the second term (17). We add row to the table, where the left column keeps halving, and the right column keeps doubling.