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A puzzle in the spirit of this puzzle. I've tried to make it a bit tougher this time, though there are no distractions of any kind. Enjoy!

enter image description here

Final answer: (4,5)

Hint 1

At the top: "Sorcery plaza"

Hint 2

In the middle: Six equations

Hint 3

At the bottom: The basic answer, based on the above.

Hint 4

In a positional number system, you need to know the base and how many are at the 0th power position, how many at the 1st power position, etc.

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  • $\begingroup$ So the top part is fairly straightforward to solve but the rest less so. Am I right to think re part1 that rot13(vg cebivqrf na beqrevat sbe gur obggbz cneg?) $\endgroup$ – Stiv Oct 28 at 17:07
  • $\begingroup$ @Stiv rot13(Abg na beqrevat ohg n onfvf sbe gur obggbz cneg). $\endgroup$ – Jens Oct 28 at 23:45
4
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The final answer is:

BASE MAGIC


The first part of the puzzle

is a magic square (or "Sorcery plaza" as the first hint puts it). Each of the black figures consists of a different number of pixels between 1 and 16, we get the following table:
$$\begin{array}{|c|c|c|c|}\hline & 13 & 12 & \\\hline 16 & & & \\\hline & & 15 & \\\hline 11 & & 1 & 14 \\\hline \end{array}$$ This can be completed to the following magic square: $$\begin{array}{|c|c|c|c|}\hline 2 & 13 & 12 & 7 \\\hline 16 & 3 & 6 & 9 \\\hline 5 & 10 & 15 & 4 \\\hline 11 & 8 & 1 & 14 \\\hline \end{array}$$


The second part of the puzzle

gives, as the hint indicates, six equations. In these equations, each of the three colours represents a number, and each equation gives the result of multiplying two of these numbers (so for example the last equation yields WHITE*YELLOW = YELLOW). If we want these three number to be different, we find that BLUE is 0 and WHITE is 1. Now YELLOW can be any number, but it seems logical to take YELLOW to be 2 (and later this will appear to be correct).


Then finally the third part of the puzzle:

The fourth hint indicates that we probably need to do something with a positional number system, but then the question arises: which base do we use? Luckily, there is also a third hint indicating that this last part should be BASEd on the above. So we let each square represent a number in the base at the corresponding position in the magic square.

Using the second part, we find that the digit at position 0 should be the number of blue pixels, the digit at position 1 is the number of white pixels, and the digit at position 2 is the number of yellow pixels. This yields the following numbers (the subscript denotes the base): $$\begin{array}{|c|c|c|c|}\hline 10_2 & & & 1_7 \\\hline & 201_3 & 5_6 & 14_9 \\\hline 1_5 & 7_{10} & & 21_4 \\\hline & 3_8 & & \\\hline \end{array}$$ This can be translated to decimal: $$\begin{array}{|c|c|c|c|}\hline 2 & & & 1 \\\hline & 19 & 5 & 13 \\\hline 1 & 7 & & 9 \\\hline & 3 & & \\\hline \end{array}$$ And finally we can translate this to get the following letters: $$\begin{array}{|c|c|c|c|}\hline B & & & A \\\hline & S & E & M \\\hline A & G & & I \\\hline & C & & \\\hline \end{array}$$ Giving the solution BASE MAGIC, which is an excellent description of this puzzle!

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  • $\begingroup$ Well done! I was beginning to give up hope that it would be solved. Your solution description is also excellent! $\endgroup$ – Jens Oct 31 at 23:08

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