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For every integer $n$ with $i$ prime factors associate a unique tile in $\mathbb{R}^m$ with $m \ge i$ as such, for every prime factor $p_j$ of $n$, the tile is a cuboid of dimension $m$ with a sidelength of $p_j$, the rest $m-j$ sides have length 1. E.g. if $m=5$ and $n=6$ the tile is a $2 \times 3 \times 1 \times 1 \times 1$ hypercuboid.

Is it possible to tile every $\mathbb{R}^m$ using each tile of prime factors less than $m$ exactly once?

Is it possible to tile $\mathbb{R}^\infty$ using the unique cuboid associated to every natural integer exactly once?

Is there any tiling of $\mathbb{R}^m$ which do not consist of infinite columns of $\mathbb{R}^{m-1} \times 1$, $\mathbb{R}^{m-1} \times 2$, $\mathbb{R}^{m-1} \times 3 \dots$ ?

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    @Optimus: I assume, seeing the answer below, you don't want to delete the question now. I$f$ that's incorrect, and you do want the question deleted, you can @-notify me.2011-10-08

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Here is the trivial tiling:

In R^2 put all nx1 tiles on a line for all n, all mx2 tiles one a line for m>=2 (to form a (inf)x2 column), all bx3 on a line for b>=3 etc. In R^3 take the tiling for R^2 which is R^2x1, then stack on top of it R^2x2, then under it R^2x3, then over it R^2x4 etc.

For any R^m use the same process with R^m-1.