We consider the fundamental problem of multiple stations competing to transmit on a multiple access channel (MAC). We are given $n$ stations out of which at most $d$ are active and intend to transmit a message to other stations using MAC. All stations are assumed to be synchronized according to a time clock. If $l$ stations node transmit in the same round, then the MAC provides the feedback whether $l=0$, $l=2$ (collision occurred) or $l=1$. When $l=1$, then a single station is indeed able to successfully transmit a message, which is received by all other nodes. For the above problem the active stations have to schedule their transmissions so that they can singly, transmit their messages on MAC, based only on the feedback received from the MAC in previous round. For the above problem it was shown in [Greenberg, Winograd, {\em A Lower bound on the Time Needed in the Worst Case to Resolve Conflicts Deterministically in Multiple Access Channels}, Journal of ACM 1985] that every deterministic adaptive algorithm should take $Ω(d (\lg n)/(\lg d))$ rounds in the worst case. The fastest known deterministic adaptive algorithm requires $O(d \lg n)$ rounds. The gap between the upper and lower bound is $O(\lg d)$ round. It is substantial for most values of $d$: When $d = $ constant and $d \in O(n^ε)$ (for any constant $ε\leq 1$, the lower bound is respectively $O(\lg n)$ and O(n), which is trivial in both cases. Nevertheless, the above lower bound is interesting indeed when $d \in$ poly($\lg n$). In this work, we present a novel counting argument to prove a tight lower bound of $Ω(d \lg n)$ rounds for all deterministic, adaptive algorithms, closing this long standing open question.}