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Sudoku

I am having problems solving this puzzle. The sum of each 3x3 square should be 2019, how to find the number in the bottom right corner? Labelling each field we can gain some information about the numbers by subtracting two neighbouring squares, but is looks like a very cumbersome and long proces. And am not even sure that it would solve the problem. Maybe one can find some invariant to use?

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Note that the sum of the top six elements in the first column is the same as the sum of the top six elements in the last column because you can make four $3 \times 3$ squares up against the top left and get a sum of $4\cdot 2019$ and four $3 \times 3$ squares up agains the top right and also get a sum of $4 \cdot 2019$. A similar argument says the sum of the bottom six elements in the first column is the same as the sum of the bottom six elements of the last column.

Now adding the top six of the first and the bottom six of the last gives the same sum as the top six of the last and the bottom six of the first. The center five squares on each edge cancel out and we are left with $10+?=11+6, ?=7$

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  • $\begingroup$ Thank you! That was fast! $\endgroup$ – Nikolaj K Nov 29 '18 at 19:20
  • $\begingroup$ @ChristianBlatter The usage of "Top N" to mean "Elements 1 through N" was first popularized by Professor D. Letterman. $\endgroup$ – MartianInvader Nov 30 '18 at 0:07
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Please consider this as a supplement to Ross Millikan excellent answer...

A sudoku like puzzle

$$(10 + ?) - (6+11) = 4(2019) + 4(2019) - 4(2019) - 4(2019) = 0\\ \implies ? = 7$$

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  • $\begingroup$ Thanks. I couldn't think of an easy way to draw the figure. $\endgroup$ – Ross Millikan Nov 29 '18 at 20:25
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enter image description here

Notice that because every $3\times 3$ square has the same sum that

$10 + A+ B = 8+E+F$ and that $A+B + 7 = E+F + X$.

Which means $X = 5$.

Likewise $7+C+D= X + G+H = 5+G+H$ and $C+D+6 = G+H+Z$

Which means $Z = 4$.

And so on.

$8 + E+F = 11 + J+K$ while $E+F+5 = J+K + Y$ so $Y= 8$.

And $5+G+H = 8 + L+M$ while $G+H+4 = L + M +?$ so $?=7$.

And that's that.

Or simply, by $3\times 3$ squares adding to the same, the difference between any two squares $3$ terms apart, must be equal to the corresponding difference of two squares $3$ terms apart in a row or column three rows or columns away.

So $10 -8 = 7 - X= 6-Z$ and 11-8 = Y-X= ? - Z$ and so.....$X,Y,Z, ?$ are easily solved.

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