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In the grid shown, we need to fill in the squares with numbers so that the number in every square, except for the corner ones, is the average of its neighbours. The edge squares have three neighbours, the others four.

grid

What is the value of the number in the square marked x?

My way is too hard: I set pronumerals a,b,c,d,e,f,g,h, and find all other cells expressions about x,a,b,c,d,e,f,g,h. for example, the top row, second column is (a+x+1000)/3, etc. Then I have Multivariate linear equations. It is too hard to solve it.

grid solve

Any one has a better method?

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  • $\begingroup$ By symmetry, $a=c=d=e=h=0$, $g=x$, and $b=f=-x$. Now try it! $\endgroup$
    – TonyK
    Commented Feb 9, 2020 at 10:37

2 Answers 2

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Symmetry might suggest

 1000    .  0    .  -1000
  .      x  .   -x     .
  0      .  0    .     0
  .     -x  .    x     .
-1000    .  0    .   1000

and then

1000     y  0   -y  -1000
  y      x  0   -x    -y
  0      0  0    0     0
 -y     -x  0    x     y
-1000   -y  0    y   1000

in which case $y=\frac13(1000+x+0)$ and $x=\frac14(y+y+0+0)$

so $x=200$ and $y=400$

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Put the origin in the centre and consider the function $xij$, where $i$ is the row index and $j$ the column index (counted from the centre). (Edit in response to the comments: This function specifies the values in the empty squares; it doesn’t apply to the values in the corners that are already filled in.) This has the desired properties. The condition for the squares next to the corners yields

$$ \frac13(1000+0+x)=2x $$

and thus $x=200$.

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  • $\begingroup$ I don't understand your $xij$. It can't be simple multiplication. It looks like you've left out the crucial definition. $\endgroup$
    – TonyK
    Commented Feb 9, 2020 at 16:30
  • $\begingroup$ @TonyK: Why not multiplication? (Note that $x$ is the unknown value, not an $x$ coordinate.) $\endgroup$
    – joriki
    Commented Feb 9, 2020 at 16:40
  • $\begingroup$ Because that would give $4x=1000$, so $x=250$, which doesn't work. $\endgroup$
    – TonyK
    Commented Feb 9, 2020 at 17:20
  • $\begingroup$ @TonyK: Ah, I see. I should have said that the function is only meant for the missing squares, not for the $1000$ that's already there :-) I've added that. $\endgroup$
    – joriki
    Commented Feb 9, 2020 at 17:39

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