Covering of a surface of a cube $n\times n \times n$ by pieces of paper $1\times 6$

Question

When I was too young one of my problems was in the list of problems of All-Russian Olympiad. The problem is the following:

Problem. We have a surface of a cube $n\times n \times n$ such that each face is divided in $n^2$ unit squares. Then this surface was covered (once) by $3n^2$ pieces of paper $1\times 2$ (domino) without overlapping. Prove that if $n$ is odd than the number of bent "dominos" is also odd.

Comment. This problem is not very difficult therefore it was solved by 90% participants in 10th grade (about 63 of 70). The solution you can find here. You really need it?)

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"Hm..." - I thought - "What will be with $1\times 6$?". It was obvious for myself that it is impossible to cover the surface for $n=1$, but possible for $n=2, 3, 4$... $n=6$...

Aga, if $n$ is even than we can divide faces in two groups (in each group $3$ faces) in such way that faces in one group forms on surface of the cube rectangle $3n\times n$ with $2$ "bends". Each rectangle can be easily divided in pieces of paper $1\times 6$.

The case when $3$ is a divisor of $n$ I am leaving for a reader.

After that I conjectured the following.

Conjecture. For $n$ such that $(n, 6)=1$ there is no such covering of the surface of the cube $n\times n \times n$ by $n^2$ pieces of paper $1\times 6$ without overlapping.

Then I prove that it is true for $n=5$ and some my friends found a more or less suitable proof but it does not work in general case (they understood that vertices of this cube are weak points in this problem).

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Finally, this conjecture was disproved (for $n=11$) by Peter Mueller(see his answer below)!

Also I like to add one problem (Russian olympiad folklore):

Problem. Let $n$ be such natural number that $(n,3)=1$, then there is no covering of three faces of a cube $n\times n \times n$ with a common vertex (of this cube) by strips $1\times 3$.

Answer 1

Indeed, there is no solution for $n=5$, and also none for $n=7$. However, for $n=11$ there is a tiling of the requested form. I found it using a straightforward exact cover formulation and Knuth's original exact cover solver. I'm not sure how to best visualize a solution, here is an attempt using random colors for the tiles. It is a little hard to check that all tiles continue correctly across the cut edges of the cube:

Added later: Fedja suggested in his comment to more generally look at $1\times 6$ tilings of a $m\times n\times k$ cuboid, for it is easy to see that such a tiling yields a tiling of a $(m+6)\times n\times k$ cuboid.

Indeed, the covering of the $11\times11\times11$ cuboid can be obtained from the way faster to compute covering of the $5\times11\times11$ cuboid:

As to the $13\times13\times13$ case, the programs (Knuth's cover and alternatively integer linear program solvers) are still running. There are no solutions for $13\times7\times7$ and $13\times13\times1$. I don't know yet about $13\times13\times7$ and $13\times13\times13$.