That "2435 isn't a multiple of 13" is the point! If it were, there would be no answer to this problem. "Modulus" problems are all about remainders. You are trying to find a number, n, such that some multiple of 2436, divided by 13n, has remainder 1, that is, such that [itex]13n= 1+ 2436k[/itex] for some integer k. That is the same as [itex]13n- 2436k= 1[/itex].
Euclidean division algorithm:
13 divides into 2436 187 times with remainder 5- that is, 2436- (187)(13)= 5.
5 divides into 13 twice with remainder 3- that is, 13- (2)(5)= 3.
3 divides into 5 once with remainder 2- that is 5- 1(3)= 2.
2 divides into 3 once with remainder 1- that is, 3- 1(2)= 1.
Replace that "2" by 5- 1(3) from the previous equation: 3- 1(5- 1(3))= 2(3)- 5= 1. Replace the "3" in that with 13- 2(5): 2(13- 2(5))- 5= 2(13)- 5(5)= 1.
Replace the "5" in that by 2436- 187(13): 2(13)- 5(2436- 187(13))= 937(13)- 5(2436)= 1.
Now that I used 2436 rather than the mistaken 2435, I get the same result as Mensenator.
