Let $\lambda$ be a partition. Suppose that $\lambda$ is both $2$ and $3$decomposable, in the sense that $\lambda$ admits a total decomposition by both $2$rim hooks (aka dominos) and $3$rim hooks. Equivalently, assume that the $2$core and $3$core of $\lambda$ is zero. Then what can be said about the $6$core of $\lambda$, if anything? Has this problem been studied (i.e. some sort of fundamental theorem of arithmetic for tableaux)? Must the $6$core either be a $2\times 3$ or a $3\times 2$ rectangle? The problem is easily reduced to the classification of $6$cores whose $2$and $3$cores are empty; I had difficulty writing down such a core other than the above two examples.
Let me comment a bit further on the fact that there are other 6cores with trivial 2 and 3core, in fact infinitely many of them. The fundamental paper of Garvan, Stanton, Kim, "Cranks and Tcores", gives a bijection between tcores and integer tuples $(n_0,n_1,\dots,n_{t1})$ which satisfy $\sum_{i=0}^{t1}n_i=0$. Under this bijection the corresponding partition has size $$\frac{t}{2}(n_0^2+\cdots +n_{t1}^2)+n_1+2n_2+\cdots+(t1)n_{t1}.$$ Now, this means that our set of 6cores can be identified with 6tuples $(n_0,n_1,\cdots,n_5)$ which satisfy $\sum_{i=0}^5 n_i=0$. In order to have an empty 2core they must further satisfy $n_0+n_2+n_4=n_1+n_3+n_5=0$, and in order to have empty 3core they must satisfy $n_0+n_3=n_1+n_4=n_2+n_5=0$. Therefore we are left with a sublattice of $\mathbb Z^6$ of dimension 2 generated by $(1,1,0,1,1,0)$ and $(0,1,1,0,1,1)$. These correspond to your two partitions, $2^3, 3^2$. Thus, we can form the generating function for 6cores with empty 2 and 3 core: $$\sum_{m,n\in \mathbb Z^2}q^{12(m^2+mn+n^2)6(m+n)}$$ This ends up being expressible as an infinite product, or a ratio of eta functions, if you will, as follows $$\frac{\prod_{k\geq 1}(1q^{12k})^3(1q^{18k})^2}{\prod_{k\geq 1}(1q^{6k})^2(1q^{36k})}.$$ Presumably generating functions of partitions that are $n$cores but have empty $d$core for all $dn$ (I would be tempted to call these "pure ncores") also have infinite product expansions, although I haven't seen this written anywhere.

1$\begingroup$ Thank you very much; this is extremely useful for what I had in mind. $\endgroup$ – Philip Engel Mar 2 '17 at 17:58

$\begingroup$ Very nice answer. I think the (6,2,2,2) partition I mentioned below corresponds under this bijection to (1,0,1,1,0,1), one of the differences between the two generating 6tuples you gave. (The conjugate, (4,4,1,1,1,1), corresponds to the other difference, (1,0,1,1,0,1), consistent with the paper's result on conjugation and these tuples.) Glad for a reason to work through some examples of their bijection. $\endgroup$ – Brian Hopkins Mar 4 '17 at 5:03
I believe a partition of 12 answers your 3rd question in the negative: Consider (6,2,2,2). Each hook length is in the set $\{1,2,3,4,8,9\}$, so it is 6core. One can check that the partition admits total decompositions into both 2 and 3rim hooks.
No answers for your more interesting questions. Initially I thought of Jaclyn Anderson's 2002 Discrete Mathematics article "Partitions that are simultaneously $t_1$ and $t_2$core" and the work it inspired, but your question is at the opposite extreme by asking about $t_1$ and $t_2$decomposable partitions.