In this paper, we consider the problem of partitioning active users in a manner that facilitates multi-access without collision. The setting is of a noisy, synchronous, Boolean, multi-access channel where $K$ active users (out of a total of $N$ users) seek to access. A solution to the partition problem places each of the $N$ users in one of $K$ groups (or blocks) such that no two active nodes are in the same block. We consider a simple, but non-trivial and illustrative case of $K=2$ active users and study the number of steps $T$ used to solve the partition problem. By random coding and a suboptimal decoding scheme, we show that for any $T\geq (C_1 +ξ_1)\log N$, where $C_1$ and $ξ_1$ are positive constants (independent of $N$), and $ξ_1$ can be arbitrary small, the partition problem can be solved with error probability $P_e^{(N)} \to 0$, for large $N$. Under the same scheme, we also bound $T$ from the other direction, establishing that, for any $T \leq (C_2 - ξ_2) \log N$, the error probability $P_e^{(N)} \to 1$ for large $N$; again $C_2$ and $ξ_2$ are constants and $ξ_2$ can be arbitrarily small. These bounds on the number of steps are lower than the tight achievable lower-bound in terms of $T \geq (C_g +ξ)\log N $ for group testing (in which all active users are identified, rather than just partitioned). Thus, partitioning may prove to be a more efficient approach for multi-access than group testing.