Let the number $1$ label the first Monday. Then the professors' schedules mean that their lecturing days satisfy the congruences below:
In order to find the Sunday in which the last professor is forced to refrain lecturing, we will take the minimum positive solution to each system $d_i=i\pmod\circ$, $d_i=0\pmod 7$ (so that professor $i$ should be lecturing by the schedule but can't because it's Sunday) and find the maximum among them. I just checked multiples of $7$ mentally until I found solutions to each. The solutions are $7,14,7,7,35,21$, so by $35$ days every professor will have omitted a lecture at some point.
On the other hand, if you're looking for the first Sunday in which all six professors simultaneously omit a lecture, then we must solve all of the above congruences for a single $d$ at the same time, with the additional congruence of $0\bmod7$. Note that the modulo $1$ congruence is trivially satisfied and can be thrown out, the modulo $2$ congruence gets subsumed into the modulo $4$ congruence and the modulo $3$ congruence gets subsumed into the modulo $6$ congruence, so reducing gives $d\equiv$
The first two can be solved into $11\bmod12$, leaving the remaining modulus's coprime so that you can use the general construction method from Wikipedia's article on the Chinese remainder theorem in order to solve this system. The computation is
hence our solution is day 371. (But what crazy classes take over a year nonstop?)