This puzzle is closely based on this one: Generating numbers with cubes

Now we want to generate Roman numerals by placing up to three 6-sided dice side by side. We are allowed to write multiple numerals on each side of a dice, such as "III" for 3 or "XX" for 20.

What is the largest contiguous range of positive numbers you can make?

Remember that:

I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1000

If you need larger numerals, then feel free to use this notation. Good luck!

UPDATE: I made a Java program that you can use to check your score. Simply pass your 3 dice as Strings (space or comma separated) to the score() method. Note that currently it does not support the flipping operation, but perhaps someone can add that.

Stealthy update from @Bass: I also made a program for checking the dice, this one's in Python, and you can try it online.

  • 6
    $\begingroup$ If I make a die with IX, can I flip it for XI ? $\endgroup$
    – Gloweye
    Commented Sep 27, 2019 at 14:38
  • 2
    $\begingroup$ Is it all three dice side by side, or up to three dice side by side? The latter allows 4 (IV) to be written whereas the former does not. $\endgroup$
    – spyr03
    Commented Sep 27, 2019 at 17:02
  • 3
    $\begingroup$ May we choose to represent 4 as either IIII or IV? $\endgroup$ Commented Sep 27, 2019 at 18:44
  • $\begingroup$ @David 4 has to be IV. $\endgroup$ Commented Sep 27, 2019 at 22:38
  • 1
    $\begingroup$ @Bass that's a very clever scorer! Thank you for your contribution. $\endgroup$ Commented Sep 29, 2019 at 22:45

5 Answers 5


UPDATE (after a pretty sturdy hint from OP):

Found a 170:


(Again, one more number can be constructed, if flipping the dice is allowed.)

Original answer:

I got all the way up to


with these dice:

 1: I  II III IV  IX C
 2: V  X  XV  XCV CX C
 3: XX L  LV  LXX C  CXX

Here's how to construct the numbers

 I              = 1
 II             = 1
 III            = 1
 IV             = 1
 V              = 2
 VI    = V.I    = 2+1
 VII   = V.II   = 2+1
 VIII  = V.III  = 2+1
 IX             = 1
 X              = 2
 XI    = X.I    = 2+1
 XII   = X.II   = 2+1
 XIII  = X.III  = 2+1
 XIV   = X.IV   = 2+1
 XV             = 2
 XVI   = XV.I   = 2+1
 XVII  = XV.II  = 2+1
 XVIII = XV.III = 2+1
 XIX   = X.IX   = 2+1

 In short: I to XIX can be made with dice 1 and 2.

 XX to XXXIX = XX(3) + I to XIX (1 and 2)

 XL     = X.L      = 2+3
 XLI    = X.L.I    = 2+3+1
 XLII   = X.L.II   = 2+3+1
 XLIII  = X.L.III  = 2+3+1
 XLIV   = X.L.IV   = 2+3+1
 XLV    = X.LV     = 2+3
 XLVI   = X.LV.I   = 2+3+1
 XLVII  = X.LV.II  = 2+3+1
 XLVIII = X.LV.III = 2+3+1
 XLIX   = X.L.IX   = 2+3+1

 L to LXIX = L (3) + I to XIX (2+1)

 LXX to LXXXIX = LXX (3) + I to XIX (2+1)
 XC     = X.C     = 2+3
 XCI    = X.C.I   = 2+3+1
 XCII   = X.C.II  = 2+3+1
 XCIII  = X.C.III = 2+3+1
 XCIV   = X.C.IV  = 2+3+1
 XCV              = 2
 XCVI   = XCV.I   = 2+1
 XCVII  = XCV.II  = 2+1
 XCIX   = X.C.IX  = 2+3+1
 C to CXIX = C(3) + I to XIX (2+1)

 CXX to CXXXIX = CXX (3) + I to XIX (2+1)
 CXL to CXLIX: like XL to XLIX, but CX on die 2 instead of X

 CL                = 2+3
 CLI    = C.L.I    = 2+3+1 
 CLII   = C.L.II   = 2+3+1 
 CLIII  = C.L.III  = 2+3+1 
 CLIV   = C.L.IV   = 2+3+1 
 CLV    = C.LV     = 2+3
 CLVI   = C.LV.I   = 2+3+1 
 CLVII  = C.LV.II  = 2+3+1 
 CLVIII = C.LV.III = 2+3+1 
 CLIX   = C.L.IX   = 2+3+1 

 and finally
 CLX = C.L.X = 1+3+2 (This could also be its own number on the sixth side of the first die.)

If I can flip the dice, there's also

161 = CLXI = C.L.XI = 2 + 3 + flipped 1

  • 2
    $\begingroup$ I can confirm that this gets to 160 without the flip and 161 with it! Great job! $\endgroup$
    – Dr Xorile
    Commented Sep 28, 2019 at 2:18
  • 2
    $\begingroup$ Thanks for the proofread, @DrXorile! When trying to verify one's own ideas, even with everything written out, it's really hard to be certain there isn't a mistake lurking in a blind spot somewhere. $\endgroup$
    – Bass
    Commented Sep 28, 2019 at 2:36
  • $\begingroup$ @Bass this is currently the best answer. However I cannot accept it just yet as one can still do a little bit better. As a hint, consider using other combinations with C, instead of using C three times. $\endgroup$ Commented Sep 29, 2019 at 23:59
  • $\begingroup$ Congratulations my friend you found the answer! This is what I found with my hill-climbing algorithm and I believe it is the optimal. $\endgroup$ Commented Oct 1, 2019 at 0:43

Partial answer, which goes up to



1. Consider the following 2 dice: I,II,III,IV,VI,VII and X,XX,XXX,XL,LX,LXX. These can display any 2-digit number whose digits are in the set of {0,1,2,3,4,6,7} (0 = just don't place the corresponding dice).
2. Let's add L and V to the 1st and 2nd faces of dice 3 (leaving the others blank). This allows to additionally express any 2-digit number which has a 5 it it (except 55). By combining V and III, or L and XXX, you can also get a number which has an 8 in it (except 58, 85, 88).
3. Add IX to the 3rd face of dice 3. This can express any number ending with 9, except 89 (note that 59 can be expressed by using L+I+X).
4. Add LV to the 4th face of dice 3. This allows to express 55 (just as LV) and 58 (as LV+III).
5. Add XV to the 5th face of dice 3. This allows to express 85 (as LXX+XV) and 88 (as LXX+XV+III). 6. To get 89, you can for example add XIX to the remaining 6th face of dice 3 (getting LXX+XIX).

So, the resulting dice are:



We still cannot get 90 or more in this configuration because it requires new C numeral (90=XC, 91=XCI etc.)

  • $\begingroup$ I wonder why are you repeating the same numerals on different dice? $\endgroup$ Commented Sep 27, 2019 at 7:04
  • 2
    $\begingroup$ Because different numerals can occur simultaneously in a number (this is radically different from Arabic numerals). For example, if we remove III from dice 3, we cannot get 58=LVIII anymore (since to use both L and V, we need dice 1 and 2). $\endgroup$
    – trolley813
    Commented Sep 27, 2019 at 7:13
  • $\begingroup$ oh I see now. This puzzle is more complicated than I thought! Perhaps replace III from dice 3 with LV? $\endgroup$ Commented Sep 27, 2019 at 7:32
  • 1
    $\begingroup$ Probably yes, so I've marked my answer as partial, since I think it's very far from optimal. $\endgroup$
    – trolley813
    Commented Sep 27, 2019 at 7:39
  • 1
    $\begingroup$ @DmitryKamenetsky I've updated my answer, now it's somewhat better. $\endgroup$
    – trolley813
    Commented Sep 27, 2019 at 8:11

First, to give an upper bound on the solution:

The maximum potential is 342 (6^3+3*6^2+3*6) as this is the number of discrete representations available with three dice where not all dice need be used.

Second, the potential solution I'll be outlining here yields all numbers up to and including:


How we get there:

The first die is our ones die and will have I II III IV and IX, which also leaves one space for a maximum value that sadly won't be able to pair with I's
The second die is our tens die, but also needs a V so it looks like V X XX XXX XL which leaves a space for trickiness later, since XC will only be fully used once (Too few places to put Vs), there is no point putting it here
The third die requires a V and an L, which takes us to 89, an XC and C to get us through 149 and a CL (150) to reach 164, so V L XC C CL with one space remaining
Now for the tricky bit, CLX (160) on the third die gets us through to 174 and XXV on the second allows for combinations with both CL and CLX to get us to 189
Finally, we stick CLXC on the largely useless sixth spot from the first die because we can

Final Dice are:


Compared to the upper bound, this solution rates a:


EDIT: I just noticed that CXV (155) is not possible with this solution

replacing CLXC with CXLV, XXV with XV, CL with CX and CLX with CXX allows us to still hit 145, for a new percentage of 42.40%

  • $\begingroup$ Nice work! I think upper bound is higher because for 3 dice we can also vary their order in 6 ways. So upper bound is 6*6^3+3*6^2+3*6=1422. It is actually greater since we can flip some of the terms. $\endgroup$ Commented Sep 27, 2019 at 13:15
  • 1
    $\begingroup$ I think it should be 1530. I believe you have more choices for two dice. My reasoning is 6 * 6 ^ 3 choices if all 3 are used (3 * 2 choices for the order of the dice, then 6 choices for each of the 3 dice on what numeral to display), 6 * 6 ^ 2 choices to use 2 dice (3 choices for the first, 2 for the second, 6 ^ 2 possible numerals) and finally 3 * 6 for a singular die. $\endgroup$
    – spyr03
    Commented Sep 27, 2019 at 17:45
  • 4
    $\begingroup$ How would you make 65 with your proposed solution? $\endgroup$
    – Dr Xorile
    Commented Sep 27, 2019 at 17:57
  • $\begingroup$ You are right - there are more choices for two dice. $\endgroup$ Commented Sep 27, 2019 at 23:39
  • 3
    $\begingroup$ It would probably be a good idea to completely rewrite this answer. As it stands, it's almost unreadable because the actual final dice aren't listed, the claimed result is later retracted, there are a lot of errors (CXV isn't 155, CLXC isn't a valid roman number), and the percentages are calculated from an incorrect theoretical limit. Also, the listed dice cannot make LXV; and while the modifications in the edit fix this, they also make LXXV impossible in the process. $\endgroup$
    – Bass
    Commented Sep 28, 2019 at 2:02

I've found a handful of solutions that will get us up to 89:

89: [['I', 'II', 'III', 'IV', 'VI', 'VII'], ['X', 'XX', 'XXX', 'XL', 'LX', 'LXX'], ['V', 'IX', 'XI', 'XV', 'XIX', 'L', 'LV']]
89: [['I', 'II', 'IV', 'IX', 'X', 'XI', 'XXV'], ['III', 'V', 'VI', 'VII', 'VIII', 'X'], ['IIX', 'XII', 'XX', 'L', 'LX', 'LXX', 'LXXX']]
89: [['I', 'II', 'IV', 'IX', 'X', 'XI', 'XXV'], ['III', 'V', 'VI', 'VII', 'VIII', 'X'], ['II', 'IIX', 'XII', 'XX', 'L', 'LX', 'LXXX']]
89: [['I', 'II', 'III', 'V', 'VIII', 'X'], ['V', 'IX', 'XI', 'XVII', 'XX', 'L', 'LXX'], ['IV', 'VI', 'VII', 'IX', 'X', 'XI', 'LXX']]
89: [['I', 'II', 'III', 'IX', 'X', 'XI', 'XVI'], ['IV', 'V', 'XX', 'L', 'LX', 'LXX'], ['V', 'VI', 'VII', 'VIII', 'X', 'L']]

Can't find anything that beats it...


9..156 = 147 contiguous numbers

For the first die,


For the second,


For the third,

C1 L C2 LV C3 C C4 CV C5 CX C6 CL


A1B1 is 9. We can construct 10..19 using B1 or B6 plus A1..A4. Likewise 20..29, 30..39. For 40..49, we can make XL/XLV using B1C1 or B1C2 and A1..A4. 50 is C1; 51..54 is C1 plus A1..A4; 55..59 is C2 plus A1..A4. 60..89 proceeds as 10..39 with C1 prepended. 90 is B1C3; for 91..94 append A1..A4. Substitute C3 for C4 to get to 99. 100 is C3; from here 100..139 proceeds as 51..89 140 is C5B3 and append A1..A4 for 141..144, then substitute B4 to get to 149. 150 is C6 and append A1..A4. Hitherto-unused are A5 and A6

  • $\begingroup$ Nearly had a good answer, but there are several problems: XVIIII is not a valid number, it should be instead XIX (and instead of IIII it's IV) $\endgroup$
    – Belhenix
    Commented Sep 27, 2019 at 21:48

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