Six professors begin courses on Monday Tuesday Wednesday Thu

Six professors begin courses on Monday, Tuesday, Wednesday, Thursday, Friday,
and Saturday, respectively, and announce their intentions of lecturing at intervals of
2, 3, 4, 1, 6, and 5 days, respectively. The regulations of the university forbid Sunday
lectures (so that a Sunday lecture must be omitted). When first will all six professors
find themselves compelled to omit a lecture? Hint: Use the CRT.

Solution

a)

First take the number 1 and label the first Monday. Then the professors\' schedules mean that their lecturing days satisfy the below:

d11(mod2),d22(mod3),d33(mod4),d44(mod1),d55(mod6),d66(mod5)

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 di=i(mod),di=0(mod7)
(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 0mod7. 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
3mod4,5mod6,1mod5,0mod7.

The first two can be solved into 11mod12,leaving the remaining modulus\'s coprime so that you can use the general construction method on the Chinese remainder theorem to solve this system.
The computation is 11.35.(35power as -1 base as 12)+1.84.(84power as -1 base as 5)=4571371mod420

so we got the solution as day 371.