The ones I have been able to find are;
fifty = 50 = DLDMC/VLVCX
fiftyeight = 58 = DIDVLMICXV / DIDVMLICXV
ninetyfive = 95 = DLDIMCXLVI / DLDIMXCLVI / DLDIXMCLVI
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:
- fifty - fiftynine
- ninety - ninetynine
Then I went through them all and tried to make moros out of them.
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.
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.
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.
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.
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.
No triples to subtract 3, and doesn't end in II.
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.
This has no solution since there is a triple, no pair.
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.
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.
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.
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.
'n' is tripled, and the only other pair is 'e' which is unusable for elimination.
Doesn't end in II and no triples to subtract 3.
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.
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;
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
$$DLDIMCVLXI = -500 -50 -500 -1 +1000 +100 -5 +50 +10 +1 = 95$$
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.
There is a triple of e's and n's. No solution.
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.
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.
$$XVXCLIXVXC = -10 -5 -10 +100 -50 -1 -10 -5 -10 +100 = 99$$
$$XVXCLIXVXC = -10 -5 -10 +100 -1 -50 -10 -5 -10 +100 = 99$$