Consider the Hales-Jewett theorem. The $k$-dimensional version of it tells us that the combinatorial space $\mathcal{U}_{M, Λ} = \{ η\mid η: M \to Λ\}$ has, under suitable assumptions, monochromatic $k$-dimensional subspaces, where by a $k$-dimensional subspace we mean there exist a partition $\langle N_0, N_1, \cdots, N_k \rangle$ of $M$ such that $N_1, \cdots, N_k \neq \emptyset$ (but we allow $N_0$ to be empty) and some $ρ_0: N_0 \to Λ$, such that the subspace consists of those $ρ\in \mathcal{U}_{M, Λ}$ such that for $0<l<k+1, ρ\restriction N_l$ is constant and $ρ\restriction N_0= ρ_0.$ It seems natural to think it is better to have each $N_{l}, 0<l<k+1$ a singleton. However it is then impossible to always find monochromatic $k$-dimensional subspaces (for example color $η$ by $0$ if $|η^{-1}\{α\}|$ is an even number and by $1$ otherwise). But modulo restricting the sign of each $|η^{-1}\{α\}|$, we prove the parallel theorem -- whose proof is not related to the Hales-Jewett theorem. We then connect the two numbers by showing that the Hales-Jewett numbers are not too much above the present ones. This gives an alternative proof of the Hales-Jewett theorem.