This is rather cumbersome and I fully expect someone will make me look silly by posting a much simpler answer.
First, we observe that a 1
placed on one of the two middle squares of a sextet will
bind all even numbers available on that sextet including the 2
. If there is a neighbour of 1
opposite the 2
it cannot satisfy both parity constraints, therefore such a neighbour cannot exist. This rules out a 1
on these squares:
. . . . . .
. x . . x .
. x . . x .
. x . . x .
. x . . x .
. . . . . .
Now let's also rule out the remaining sextet middle positions:
Focusing on a stack of two sextets there is up to symmetry only one case
. x . x O x E O x
. x . . x . 1 E .
==> ==>
. 2 . . 2 . E 2 .
E 1 E E 1 E E 1 E
Here E
means (must be) even, O
odd and x
can't be 1
. We see that a 1
placed in the bottom middle position ultimately forces four even numbers on that sextet, contradiction.
Ruled out so far:
. x . . x .
. x . . x .
. x . . x .
. x . . x .
. x . . x .
. x . . x .
Next, we rule out the following configuration of two 1
s.
. x . E x .
1 x . 1 E E
==>
. x 1 E E 1
. x . . x E
As not both 1
s can be hugging the boundary at most one of the E
s can be a 2
, contradiction.
From this we can conclude that given a 1
in the corner of a sextet, the vertically adjacent sextet (if it exists) must have even numbers in two specific positions:
. . . E . . E . .
1 . . 1 E . 1 E .
==> ==>
x x x E x x E . .
. x . . E . 1 E .
Furthermore, none of the four indicated E
s can be a 2
.
We also observe that of the two candidate positions for the second 1
only the left one is viable, because the top neighbour of the right one cannot be a 2
but no other evens are left.
And, finally, we remark that a three-storey version of this configuration is not possible because some of the E
s have to be 4
s.
Next we zoom in on the two middle sextets:
This configuration is impossible
. x . . x .
1 x . 1 x .
||
\/
E x . E x .
1 E E 1 E .
because none of the three indicated E
s in the left sextet can be a 2
.
Please note that this argument also holds for the bottom row of sextets if the middle left sextet has its 1
in its lower half.
. x . . x .
1 x . . x 1
||
\/
4 x . . x 4
1 6 . . 6 1
In this configuration of 1
s the indicated 4
s are 4
s because they cannot be 2
s or 6
s, leaving for the other non-2
s only 6
s. But now none of the remaining squares can be a 2
, contradiction.
What we know so far:
Both the left and the right halves must have 1
s in the top and bottom rows; the respective third 1
s cannot be on the same row. Also they must be in the same column as the closer one of the bottom or top row 1
s.
Furthermore, the ones farther away can only be at the very corner.
This rules out that the closer ones are in the corner because
1 E . . . 1
E . . . . E
1 . . . . .
. . . . . 1
. . . . . .
1 E . . . 1
we remember that one of the two top left E
s has to be a 6
.
But they cannot be in the centre either
. . 1 . E 1
. . . . . E
. E 1 E . .
. . E 1 . .
. . . . . .
1 . . 1 . .
because there cannot be a 2
in the middle right sextet.
So the answer is
No, it is not solvable.