Mathematical Rebus I

Mathematical Rebus II

First Image

Second Image

Third Image

Small note on the first image:

I wish romans had a number for 0.

Hint 1 for the squares image:

It is mostly an arithmetic problem.

Hint 2 for the squares image:

Color and relative position matter.

Hint 3 for the squares image (alternative image that should help a little bit):

Image Hint

Hint 4 for the squares image:

This is a Mathematical Rebus, so what about changing the squares by numbers? Which? It's your work to discover.

  • $\begingroup$ You could avoid needing a zero by summing from I to M and then subtracting I inside the summand. $\endgroup$ – Gareth McCaughan Jun 19 '17 at 10:49
  • $\begingroup$ @GarethMcCaughan I'm aware of that. I considered simply starting at I and summing a to it all. Even though, I felt this way made it easier to solve, so I choosed to keep it this way. Plus I wanted to make sure I was using a number that's clearly a number (like CMXCIX). $\endgroup$ – Masclins Jun 19 '17 at 10:57
  • $\begingroup$ Seems the Rebus got lost without the correct answer being found :( $\endgroup$ – Masclins Jun 4 '18 at 8:16

Building on Gareth McCaughan's answer

The first part yields Ma as Gareth explained.
The second part, the squares... the small square is just a variation of the first square. The differences are the size and the color. So this can be a "thematic" difference between the 2 squares.
The third part yields A (again as Gareth says).

Put them together and you get


  • $\begingroup$ Clever but gosh, I hope that isn't the intended answer... $\endgroup$ – Gareth McCaughan Jun 19 '17 at 16:56
  • $\begingroup$ It is not. It's way more specific and mathematics-related. $\endgroup$ – Masclins Jun 20 '17 at 8:39
  • 1
    $\begingroup$ @AlbertMasclans. I thought so, but this fit very nicely and gave me a warm feeling. :) $\endgroup$ – Marius Jun 20 '17 at 8:49

The first expression

is 0 when a=0 and increases by 1 each time a increases by 0.001. So it's floor(1000a) which, given all the Roman numerals, I should probably write as $\lfloor Ma\rfloor$.

The yellow square

is about 0.61 times the size of the white one. Perhaps this is hinting at the golden ratio (often written $\varphi$ or $\tau$) but there might be some further thing I'm missing.

The fraction at the end

is probably "voltage, in volts, over resistance, in ohms", yielding "current, in amps" or $I_A$.

So perhaps we are looking at

$\lfloor Ma\rfloor\varphi I_A$ or the Mafia. Eek! (The Mafia would fit quite well with the two earlier answers of PIZZA and SECRET, of course.)

[EDITED to add:] OP indicates that I don't have the intended interpretation of the yellow square, so here is another possibility:

In WYSIWYG equation editors, you often see something very much like this (though much smaller) to indicate a subscript. E.g., this is what the relevant menu item in Word's equation editor looks like: subscript and this is what you get when you move your cursor into the subscript: selected subscript. In that case, this could indicate SUB, making our final answer MASUBIA, the name of a group of people in central Africa.

and another:

if, as above, it's denoting a subscript, perhaps we should replace it with something that commonly appears in subscript position. The most obvious things would be $i,j,k$ but none of those makes a word -- but $n$ yields MANIA, which seems at least kind possible.

I confess that neither seems terribly likely.

  • $\begingroup$ I find it really funny that you got the that letter which would sound like that. Even though, that part is not correct. If their ratio is really near that must be because doing it manually made them look more beautiful. $\endgroup$ – Masclins Jun 19 '17 at 11:02
  • $\begingroup$ I'd point out that we have k incrementing in the summation, not a. So if the M in the summation is 1000, we have floor(a+x), where x=k/M and is always less than 1, so we're really just summing a 1000 times. (This might be different than the original, given the last edit time.) Same potential result but different reasoning. $\endgroup$ – Duncan Jun 19 '17 at 19:21
  • 1
    $\begingroup$ @Duncan Note that a need not be an integer. $\endgroup$ – Gareth McCaughan Jun 19 '17 at 21:31
  • $\begingroup$ Ah, an excellent point. Hate it when I make assumptions. $\endgroup$ – Duncan Jun 20 '17 at 0:59
  • $\begingroup$ @GarethMcCaughan by the way, the reason to keep the 0 was to be as faithful as possible to Hermite's identity. $\endgroup$ – Masclins Jun 20 '17 at 8:40

Since this remains unsolved, I proceed to answer the question.

Most of it had been solved, so I only put together the already given answers and add the missing clue.

First clue, by Gareth:

is 0 when a=0 and increases by 1 each time a increases by 0.001. So it's floor(1000a) which, given all the Roman numerals, I should probably write as $\lfloor Ma\rfloor$.

Second clue:

The trick is using the colors. White is #FFFFFF and yellow #FFFF00. It was intended to be FFFFFF module FFFF00, or $FFFFFF_{FFFF00}=FFFFFF \mod FFFF00 \equiv FF$

Third clue, also by Gareth:

is probably "voltage, in volts, over resistance, in ohms", yielding "current, in amps" or $I_A$.

All together, the answer is:

Maffia, as actually Gareth already got.

  • $\begingroup$ Following advices on Meta I choose my own answer as best in case somebody else wants to use this rebus in the future. Even though I'd like to note that Gareth solved most of the puzzle. $\endgroup$ – Masclins Jun 12 '18 at 15:04
  • $\begingroup$ What in the second clue indicates "modulo"? (The subscript? I don't think I've ever seen subscripts used that way before.) $\endgroup$ – Gareth McCaughan Jun 13 '18 at 2:56
  • $\begingroup$ When working on the set Z/nZ (for mod n) is not that rare to use such notation. I'm sorry you feel it was a very weird notation. I tried to point into the right direction through Hin 3. $\endgroup$ – Masclins Jun 13 '18 at 9:54
  • $\begingroup$ This link from Wikipedia shows the use I'm refering to: en.wikipedia.org/wiki/Modular_arithmetic#Integers_modulo_n $\endgroup$ – Masclins Jun 13 '18 at 9:54
  • $\begingroup$ I suppose it does. I'm a mathematician and don't recall ever seeing that notation before; I wonder whether whoever wrote that bit of the Wikipedia page just made it up. $\endgroup$ – Gareth McCaughan Jun 13 '18 at 12:23

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