10
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Recently a tablet from ancient rome was discovered containing a list of cryptic numbers. Most of the numbers are easy to read, but there are scratches over the last three making them difficult to decipher:

2
4
2

2
1
1
5
1

11
1
2
5
2

2000
10
2000
2
220

200
1
2
4
20

1100
1
110
2
????
????
????

What are the last three numbers?

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11
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ANSWER

9
1
1

Explanation...

Thanks to @charfellow because it is indeed an alphametic! ...But the title gives it away. :( I personally had to replace the numerals with unrelated roman characters so that it wouldn't interfere.

A quick lookup of 2-digit squares revealed only one that fit the pattern $MM^2 = CCXX$, which is $$88^2 = 7744$$ Following the logic from there, we only have one digit left to solve. The key is the $$XI-II \gt II$$ equation. Since we already found $X$ from the previous equation, we know that $I$ must be less than 4, or else we'd be getting into negatives. The only digit that fulfills the inequality is 1. So, in all, we have...

I = 1
X = 4
C = 7
M = 8

Using these values, we can fill in the equation set.

$$\begin{cases}11\ge11\\11-1\gt1\\41-11\gt11\\88*88=7744\\77-11\ge44\\87-74=????\end{cases}$$

The missing value is 13. Using only the digits we found with the alphametic, we must find an equation that equals 13 and make sure that the translation back into Roman numerals yields a legal value. Luckily, we find one right away with $87-74=14-1$. In Roman numeral alphametic, the new equation becomes IX-I. In plain Roman numerals, it becomes IX I I, or, 911.

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  • $\begingroup$ Can you show the steps that result in that answer? I'm getting a different answer with my steps (although your explanation is pretty solid). To be clear: it's the end of a long day for me. $\endgroup$ – charfellow Mar 17 '16 at 23:09
  • $\begingroup$ Working on it. Just wanted to get an answer up. $\endgroup$ – feelinferrety Mar 17 '16 at 23:15
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    $\begingroup$ It's up. I was new to MathJax. :\ $\endgroup$ – feelinferrety Mar 17 '16 at 23:37
  • $\begingroup$ Nice job, I was worried I had made it too hard but not for you! $\endgroup$ – Joel Mar 18 '16 at 3:25
  • $\begingroup$ This answer only makes sense if somebody in ancient Rome knew positional number systems before they were invented. $\endgroup$ – Stig Hemmer Mar 18 '16 at 9:17
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Not an answer

I doubt that an ancient Roman tablet would have numbers in Hindu-Arabic, so I've translated the numbers to Roman Numerals, in case it helps someone else solve the puzzle.

II
IV
II

II
I
I
V
I

XI
I
II
V
II

MM
X
MM
II
CCXX

CC
I
II
IV
XX

MC
I
CX
II
????
????
????

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    $\begingroup$ Came here to do that. You beat me to it. :P $\endgroup$ – feelinferrety Mar 17 '16 at 19:47
  • $\begingroup$ You had time - I triple checked myself :P $\endgroup$ – Khale_Kitha Mar 17 '16 at 19:48
  • $\begingroup$ I was responding to other notifications and just had the translation sitting around saved in my URL bar. Plus, I'd put everything in horizontally without spaces until a line break. Eh, I just went ahead and monospaced everything for the sake of alignment in case it matters. $\endgroup$ – feelinferrety Mar 17 '16 at 19:49
  • $\begingroup$ Nice, I didn't know you could monospace. Thanks! $\endgroup$ – Khale_Kitha Mar 17 '16 at 19:53
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    $\begingroup$ Technically, they're in code blocks `xyz` which are usually gray but the nature of the spoiler box keeps 'em yellow. ;) $\endgroup$ – feelinferrety Mar 17 '16 at 19:55
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Based on @Khale_Kitha's partial answer...

Alphametics?
Let's turn things on their sides...
$$II \ge II$$
$$II - I > I$$
$$XI - II > II$$
$$MM x MM = CCXX$$
$$CC - II \ge XX$$
$$MC - CX = ????$$
$$????$$
$$????$$
So the last line seems like it could be 1100, 1, and 110 (MC - CX = MC - CX?)

Solving time!

The only valid value for $M$ is $8$; $C$ is $7$ and $X$ is $4$
Because $$77 - 44 \ge II,$$
I = $1$ or $2$ or $3$
Because $XI - II > II$, $X > 2*I$, so $I = 1$
Last line: $MC - CX = 87 - 74 = 13$
The second number must be a plus or minus, but we have no minus, so #2 is $I$ (1)
The last line must be 87 - 74 or 84 - 71
The 3 numbers are $MC, I, CX$ (1100 1 110) or $MX, I, CI$ (1010 1 11)

All of that said, I like feelin's answer, and I hope I made a mistake! :P

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  • $\begingroup$ I wouldn't have thought of that. Nice. $\endgroup$ – Khale_Kitha Mar 17 '16 at 21:59
  • $\begingroup$ Looks like I X C and M have alternate values. Might be worth trying replacing with numbers. X > I, C > I + X, M >= 3 $\endgroup$ – Z. Dailey Mar 17 '16 at 22:54
  • $\begingroup$ Also M > C. I'm heading to dinner but if it's not solved when I get back I'll give it a go. $\endgroup$ – Z. Dailey Mar 17 '16 at 23:00
  • $\begingroup$ I have to say, this was an absolute revelation! Total genius on finding the proper format. Nice work! $\endgroup$ – feelinferrety Mar 17 '16 at 23:37
  • $\begingroup$ @feelinferrety I don't get credit for the substitution? $\endgroup$ – Z. Dailey Mar 18 '16 at 0:39
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Based on @charfellow's answer...

It could be...

9
9
0

"990" is what you get when you subtract 110 from 1100.

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  • $\begingroup$ That doesn't exactly follow the format proposed. $\endgroup$ – feelinferrety Mar 17 '16 at 22:18
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Not an Answer: Alternate Formatting

Single line breaks -> spaces, double line breaks -> new row:

II IV II
II I I V I
XI I II V II
MM X MM II CCXX
CC I II IV XX
MC I CX II ???? ???? ????

OR

Only double line breaks are preserved:

IIIVII
IIIIVI
XIIIIVII
MMXMMIICCXX
CCIIIIVXX
MCICXII????????????

OR

Order of placement in each individual list is maintained but lists are otherwise combined (AKA rows are kept but column is reduced to 1):

IIIIXIMMCCMC
IVIIIXII
IIIIMMIICX
VVIIIVII
IIICCXXXX????
????
????

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