Here come mo -Roman numerals — just like Roman numerals, only mo’ so.   (Explained as we go.)

What number can spell itself in two different ways with mo-Roman letter substitution?

Update:   Trenin found more than one solution, including a three(!)-way standout and another that stands out for using two different sets of digits/ letters rather than reordering a single set.

For example, the number 6 can spell itself in one way with mo-Roman letter substitution:

         6   =   VX I   =   s i x   by substituting   Vs,   Xi   and   Ix

But, 6 = VX I ?   Not the usual V I ?   Both work in this system — the mo-Roman the merrier!   Any smaller mo-Roman digit counts as negative if it is written to the left of any larger digit. Thus, when a smaller V is to the left of a larger X:
                  VX I   =   (−V)+X+I   =   (−5)+10+1   =   6

Self-spelling via mo-Roman substitution also works for the number 19.

   19   =   L I L XCXXL   =   n i n e t e e n   by substituting   Ln,   Ii,   Xe   and   Ct

Note that a smaller digit is negated even when it does not touch the larger digit to its right.

L I L XCXXL   =   (−L)+(−I)+(−L)+(−X)+C+(−X)+(−X)+L   =   −50−1−50−10+100−10−10+50   =  19

(In L I L XCXXL, only C and the last L count positively as they have no larger digits after them.)

Spellings of the following numbers have few enough different letters to be candidates for this puzzle:
one two three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen seventeen eighteen nineteen twenty twentyone twentytwo twentythree twentyseven twentynine thirty thirtytwo thirtythree thirtysix thirtyeight thirtynine forty fortyone fortytwo fortythree fortyfour fifty fiftyone fiftytwo fiftythree fiftyfour fiftyfive fiftysix fiftyeight fiftynine sixty sixtytwo sixtysix sixtynine seventy seventyone seventyseven seventynine eighty eightythree eightyeight eightynine ninety ninetyone ninetytwo ninetythree ninetyfive ninetysix ninetyseven ninetyeight ninetynine hundred onehundred hundredone onehundredone hundredthree hundrednine hundredten threehundred threehundredthree ninehundred ninehundrednine zero minusone minussix minusseven minusnine minusten minusnineteen

Only names of numbers (not descriptions) and one-to-one substitutions are officially in play, although variety is welcome for the sake of variety.   Also feel free to post any number that has even one way to spell itself with mo-Roman substitution (other than 6 or 19, unless you find a second way for either one).   And remember, there were in mo-Roman times.

  • $\begingroup$ one thing I don't understand. Should I try to represent the numbers you mentioned or find others? Or both? Or neither? $\endgroup$
    – Marius
    Commented Apr 21, 2017 at 8:25
  • $\begingroup$ There is a solution among the numbers listed, Marius. I'm fairly sure that no others could even work, but would love to be shown otherwise. $\endgroup$
    – humn
    Commented Apr 21, 2017 at 8:26
  • 1
    $\begingroup$ And for the record, at first glance I've read "mo-ron numbers". Which kind of fits :D $\endgroup$
    – Marius
    Commented Apr 21, 2017 at 8:26
  • $\begingroup$ Thank you, Marius! I was hoping to not be the only one to hear that in my head. $\endgroup$
    – humn
    Commented Apr 21, 2017 at 8:27
  • $\begingroup$ If I'm not misunderstanding the question, e.g. 130 can be written as XCXL or XXCL. $\endgroup$
    – Nautilus
    Commented Apr 21, 2017 at 9:00

2 Answers 2


The ones I have been able to find are;

fifty = 50 = DLDMC/VLVCX
fiftyeight = 58 = DIDVLMICXV / DIDVMLICXV
ninetynine = 99 = XVXCILXVXC / XVXCLIXVXC

I didn't have any luck with the ones over 100, nor less than zero. I did not check long number names very rigorously (e.g. onehundredone), so I might have missed some. The trick I found was getting multiple letters to offset a single letter - like 'D..D..M' or 'V..V..X'. These are obvious in the ones I found solutions for. Also, I was able to rule out some combinations due to lack or letters, or too many unique letters.

I will admit that I used a computer to verify my answers.

Also, other than the examples of 6 and nineteen, I couldn't find any other numbers that had even a single representation.

Edit to show work

Here is some of the work I did. Much of it was intuitive, so I didn't write it down explicitly, but I have tried to explain it here.

In general, many of the numbers need to include 'M/D/C', but doing so makes them too big. In some cases, you can eliminate them such as:

  • ..D..M..D.. - In this case, one D is negative and the other positive. This requires a bigger number in the middle, so doesn't really help to get rid of big numbers.
  • ..D..D..M.. - The two Ds are negative and the M is positive and is thus eliminated.
  • ..C..C..C..C..C..D.. - the five Cs are negative and the D is positive and is eliminated. Needs five of one letter though...

So the only real viable way to get rid of big numerals is with the middle option. The problem space is quite large, so I hand picked some candidates based on patterns. My criteria was:

  • must include at least one pair. Triples are not included in pairs.
  • pair must be "early" in the number. If it is too late, then you run the risk of putting the biggest numeral at the end, thus negating everything and making the number negative. For exmample, 'thirty' has a pair of 't's, but in order to use the technique above, we would need 'L...LC', which would be negative.
  • longer numbers might need to include 2 pairs to get rid of both 'DDM' and 'LLC'
  • If all numerals are used, certain numbers are impossible. For example, sixty is impossible because you can only go as high as 'C', but that means you need to use an 'I', which cannot end in 0.

Some other things to note:

  • to make a number ending in 2 or 7, you either need to subtract 3 from something, or have the number end in 'II'. Any 'I' not at the end must be negative since it is guaranteed to be followed by a bigger numeral.
  • to make a number ending in 3 or 8, you either need to subtract 2 from something, or have the number end in 'III'.

With this in mind, I started with the following candidate numbers:

  • fifteen
  • nineteen
  • thirtysix
  • fifty - fiftynine
  • ninety - ninetynine

Then I went through them all and tried to make moros out of them.

fifteen There are 5 unique numbers, so we need to include the letters 'IVXLC' or higher. We can eliminate LLC or DDM with the pair of 'f's and another letter. This other letter cannot be 'n' because that would make the whole number negative. It cannot be 'e' because there are 2 of them. Therefore, it must be 't'. This leaves us with 'e+e+n-i=15'. There is no solution for this.

nineteen This one has four unique numbers, so we will need the letters 'IVXL' or higher. The use of 'L' means we would need to subtract a lot to get back down to 19, which also puts the 'L' late in the number. But our only choices late are 'e' and 'n', both of which are multiples. Thus, it must be 't'. But to subtract enough, the 'n' would have to be X. Unfortunately, it also occurs after the 't', so there is no solution using 'L' as the highest numeral.

Instead, lets use the 'LLC' elimination. We need to do this early, so we get 'n=L' and 't=C'. Because 'L' is at the end, we need to include enough negative 'X's to counter it, so 'e=X'. Luckily, 'i=I' is also negative and gives our first solution!!

$$nineteen = LILXCXXL = -50-1-50-10+100+10+10+50=19$$

This is the only solution for nineteen.

thirtysix There are 7 unique numerals, so we need to use all 'IVXLCDM'. We need to get rid of 'DDM' and it cannot be done too late, so it must be 't=D'. We can then choose either 's' or 'y' for 'M'.

To end in 6, we need 'x=I'. Notice that IVXLC=34. Thus we need the X and L to be before the C, forcing us to have 's=C'. But then V is after the C and we are too high.

fifty There are only 4 unique numbers, so we only need to go up to 'L' if we want. However, if we do that, then we would need to offset all but the L. There is no way to do that with 3 more numerals with one of them paired.

If we go up to C, then we need to put it late in the number and get -50 somewhere. It can't be 'y' since the remaining digits are well under or well over 50, missing the mark. It can't be 'f' since that is a pair. Thus, we must have 't=C'. Since X=VV, we can eliminate the 'f's and the 'y' by making them V and X respectively and putting them on opposite sides of the C. Setting 'i=L' gives a result!.

$$VLVCX = -5-50-5+100+10=50$$

We are looking for multiple solutions, so lets keep going. Lets see if we can use the 'LLC' elimination. Thus, 'f=L', and 't' or 'y' is 'C'. But neither of these work since we can't get 50 with the remaining numerals.

So try the 'DDM' elimination instead. Thus, 'f=D'. If 'y=M', the entire number is negative, so use 't=M'. Luckily, 'y=C' and 'i=L' gets us the right result!

$$DLDMC = -500-50-500+1000+100=50$$

This gives our first solution.

fifytone There are 7 unique letters, so we need to use all the roman numerals on this one. The 'DDM' elimination is required, so 'f=D'. Our choice for M is mostly irrelevant - we have 5 unique letters to add to 51. The closest we can get is 46 or 54.

fiftytwo No triples to subtract 3, and doesn't end in II.

fiftythree The highest letter is again 'M' and we have three pairs. The 'e' is useless for elimination and must be positive. Of the 'f' and 't', one needs to be 'D' and the other 'I' to get a -2 offset from 5 down to 3. So 'y' or 'h' is M. Either way, we have '-2-i-y+r+e+e' or ,'-2-i+h+r+e+e' neither of which has a solution of 53.

fiftyfour This has no solution since there is a triple, no pair.

fiftyfive It has a pair of 'i's. We would need 'i=D' and 'v=M'. The rest of the numbers 'fffty' are negative and only 'e' is positive. We can get close to an answer by choosing 'e=C' and 'f=X' and 't=V'. But we would need 'y=15' which doesn't exist.

fiftysix The 'i' is not usable for the DDM elmination, so we need to use 'f=D'. To get a 6, we need 'x=I'. To get 6 we either need positive V, or positive X and negative V. However, this makes the assignment of 'i' L or higher, which makes everything negative.

fiftyeight Again we have 3 pairs, but the 't' cannot be used for elimination. It will, however, eliminate itself regardless of our choice for the DDM elimination because one t must be positive and the other negative.

Lets try the 'i=D' first. Thus, either 'g' or 'h' is M. We already know that 't' eliminates itself, so if 'h=M' the whole number becomes negative. Instead, use 'g=M'. We must then set 'f=I' to get our -2, and 'h=C' to get something positive to work with. Unfortunately, the L ends up negative and the whole number is less than 50.

Instead, lets try the 'f=D' elimination. We need 'i=I' and both must be negative. The first is already negative. The second will automatically be negative since it is not the last digit, so whatever comes after forces it to be negative.

We need C to be positive (thus, after the M) and L to be negative (before the M or C). We also need the V to be self eliminating (t) and the X to be positive. This forces CX ordering, but the ML ordering can be either or.

$$DIDVLMICXV = -500 -1 -500 -5 -50 +1000 -1 +100 +10 +5 = 58$$ $$DIDVMLICXV = -500 -1 -500 -5 +1000 -50 -1 +100 +10 +5 = 58$$

Gives us our second solution.

ninety There are only 5 unique digits, meaning we only need to go as high as C. But that would mean we need to use an 'I' and a 'V', and we can't eliminate both to make a nice round 90.

So lets go higher and try an DDM elimination. Thus, 'n=D'. Then we have iety to work with, with one of ety being M. Too bad it wasn't spelt ninty, because then we could go with 'DXDMC'. That extra letter messes us up and gives no solution.

ninetyone 'n' is tripled, and the only other pair is 'e' which is unusable for elimination.

ninetytwo Doesn't end in II and no triples to subtract 3.

ninetythree Need to subtract 2 to end in 3, so one of 'n' or 't' needs to be I, and the other is used in DDM elimination. Either way, the 'M' has as options 'y', 'h', or 'r'. Leaving only 'h' or 'r' or 'e' as the choice for C. Neither 'r' nor 'e' works since there is too much negative accumulation. 'h=C' forces 'f=D' and 'y=M' and 't=I'. This gives -2-i-e+r+e+e with VXL still to be assigned. The L is the one that kills us.

ninetyfive To end in 5, we need to offset the I by setting 'e=I' thus making one negative and the other positive. This forces the DDM elimination to use 'n=D'. The L will hurt us again if we can't eliminate it, so lets set 'i=L' to make one positive and the other negative.

This give us the following so far; $$DLDItyfLvI$$

We need C and X to be positive and V to be negative. V can be negative by either the M, C, or L, so there are three solutions!

V negative by M: $$DLDIVMCLXI = -500 -50 -500 -1 -5 +1000 +100 +50 +10 +1 = 95$$ V negative by C: $$DLDIMVCLXI = -500 -50 -500 -1 +1000 -5 +100 +50 +10 +1 = 95$$ V negative by L $$DLDIMCVLXI = -500 -50 -500 -1 +1000 +100 -5 +50 +10 +1 = 95$$

ninetysix Only the 'n' can be used for DDM elimination. Also, to get 6, we need 'x=I'. The L is pesky and must eliminate itself, so it must be 'i=L'. Unfortunately, this makes both the X and V negative and we can no longer get above 90.

ninetyseven There is a triple of e's and n's. No solution.

ninetyeight 4 pairs - n, e, i, and t. The 't' cannot be used for elmination. Lets try 'i=D', Thus, 'g=M'. We need to eliminate the 'L's, so 't=L' and 'h=C'. The remaining letters are all negative, making the result too log.

Instead, lets try 'e=D'. Again, 'g=M', so we end up too negative.

If 'n=D', then we must choose 'y=M'. One of 'e' or 'i' must be 'I' to get -2. t is self eliminating regardless of its value, and we need V to be paired as well. This leaves the letters 'gh' and one of the pairs, with the numerals XCL to be assigned. If L is one of the pairs and it eliminated itself, and C is positive, then X is extra and cannot be assigned.

ninetynine There are 4 n's and 2 e's. But we only need to go as high as 'C' since there are only 5 unique letters. This is ok, because the only candidates for elimination is 'i=D', and then either 'n' or 'e' is M which is impossible. Thus, we know that any solution uses all the letters IVXLC.

There are only two single letters, 't' and 'y'. One of them needs to be I to get a -1 so we can end in 9. The rest needs to sum up to 100.

There are four 'n's, two 'e's, two 'i's, and one more single. The letters to assign are VXLC. If we make the single C, then it will be positive and the rest have to cancel out. The 'L' could cancel itself, and the 'n's and 'i's too. But that would mean that two 'n's are positive and one 'i' is positive. The 'nin' at the end would need to be all positive, but there is no way to do that.

If the L is the single, then any choice of C will come after L making it negative. The pair of Cs will be 200, and take away the L is 150. 4 Xs and 2 Vs can make 50 if they are all negative, so this forces the C to be the last position, so 'e=C'. Then 'n=X' and 'i=V'. The order of LI can be switched.

$$XVXCLIXVXC = -10 -5 -10 +100 -50 -1 -10 -5 -10 +100 = 99$$ $$XVXCLIXVXC = -10 -5 -10 +100 -1 -50 -10 -5 -10 +100 = 99$$

  • $\begingroup$ This multi-solution deserves more than the current IIVX votes of approval, including a bounty (eventually, as Marius will get the first one) $\endgroup$
    – humn
    Commented Apr 22, 2017 at 19:12
  • 1
    $\begingroup$ Computer just for verifying, or actually finding as well? Regardless: good work. (I consider coding up a good search algorithm generally a very valid solution unless the puzzle requests to find a logical deduction route explicitly.) +1 (and +1 for the good puzzle idea to @humn. Still waiting for another puzzle though ;-) ) $\endgroup$
    – BmyGuest
    Commented Apr 23, 2017 at 12:29
  • $\begingroup$ What an explanation! This answer is now effectively a clinic on mo-Roman alphametic construction. $\endgroup$
    – humn
    Commented Apr 25, 2017 at 18:06

Found it:

DLDIMCXLVI = $-500 -50 -500 - 1 + 1000 + 100 -10 + 50 + 5 + 1 = 95$
DLDIMXCLVI = $-500 -50 -500 - 1 + 1000 -10 + 100 + 50 + 5 + 1 = 95$

My pseudo-strategy
I've used a "learning algorithm"
At first I took blind stabs at short words randomly, trying to use the smaller numbers IVXL That backfired quickly since the letters did not repeat themselves and could not get numbers that end with 6, 7 or 8 easily. So I ignored those.
Then I moved on to round numbers (multiples of 10) because it's easy to do the math with them. That backfired when I realized the sum of all roman numerals is 1666. So I had to NOT use the 1 (I). This lead to the use of 500 (D) or 1000 (M) which ended up being too large for the actual value of the number.
Then I started trying with numbers that end with a 6 or 9 so I can use the 1 (I) and maybe not end up using the 1000 (M).

After a few failed tries I thought of trying things that end with 5 and have the last letter duplicated somewhere in the middle of the word so I can make that letter 1 (I) because this way it will be subtracted in the middle and added at the end.
I also looked for a word that has a duplicated letter near the start so I can make that letter 500 (D) so they will be canceled out with 1000 (M) after them. Then I could re-arrange the remaining numerals in a pretty low range.

I went through the list of words (by the way, thanks for providing them so I don't have to search myself) and the number I listed as answer popped up quickly using my assumptions.

Just for fun, here are some failed attempts grouped on the categories I mentioned above:

eight, seven, nine, two
thirty, forty, fifty, sixty, hundred
sixtysix, fiftysix, fortysix, ninetysix, sixtynine (I really wished this was the one, for ...reasons)

Note: Thanks for making me waste more than 1h of my life on this for some fake internet points.
Since the title of the puzzle is "Self-intro to mo-Roman numerals" I have a feeling I will lose more hours in the future with this type of puzzles.

  • $\begingroup$ What is the second way? $\endgroup$
    – ffao
    Commented Apr 21, 2017 at 16:10
  • $\begingroup$ Second way for what? $\endgroup$
    – Marius
    Commented Apr 21, 2017 at 16:13
  • $\begingroup$ Bolded part of the question: "What number can spell itself in two different ways with mo-Roman letter substitution? " $\endgroup$
    – ffao
    Commented Apr 21, 2017 at 16:14
  • $\begingroup$ Aparently I cannot read. I will delete it. $\endgroup$
    – Marius
    Commented Apr 21, 2017 at 16:16

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