I understand how the inverse part can be a bit tricky for people to understand, since he did not explain it very clearly, so I'll show you how you can think instead. When you come to the part of 5x congruent to 2 for mod 9, imagine the list of numbers that 2 could be instead of 9. By adding 9 to 2 repeatedly, you'll get said list. The list you get is 2, 11, 20, 29, 38.... Here you see that 20 is divisible by 5, such that we can get that 5x/5 is congruent to 20/5 for mod 9, which is equal to x congruent to 4 for mod 9, and there you have your answer.
@@DavisMaths I am not sure, but this is the method that I found to be the easiest to understand. It's not that difficult if you just write out the list. I did not understand congruence until the idea of lists was mentioned in some youtube video I saw. You can think of it like this, if you have a congruent to b for mod n, the list will be {b, b+n, b+2n, b+3n, ....} This method is useful since it can also be used for squares. For example, if x^2 is congruent to 56 for mod 8, we know that you can not take the square root of 56. But if we write out the list: {56, 64, 72, 80, ...}, we find almost immediately that 64 is a square, as such we can take the square root on both sides and find the answer for x. Hopefully this explanation suffices. Goodluck!
I’m a retired life long learner who has always been interested in maths and number theory/math Olympiad problems is my current buzz. These videos are really great for bringing out the big points that might get missed with self study.
what do we do when the number 56 is raised in a very high power ? For example I was given this one: 8x ≡ 11^41 (mod 51) but I can't figure out how to solve it.
Inverse (mod 9) has an analogy to rational numbers e.g. (1/2)*2=1 so if you multiply a number by its inverse you get 1. The same situation with (modulo 9). 5*2=1 because 5*2=10 and 10 gives remainder 1 by dividing by 9. That's why 5^(-1)=2
There are. But after n=27, i.e. x=247, the next one is x=256 and that is the same as 4 (mod 252) and we go round the same set of 28 numbers differing by 9 each time. Remember we are solving 149x ≡ 56 (mod 252), so any of the infinite number of solutions will always reduce to the same set (mod 252). It is trivially true that all linear congruences (mod m) which have a solution x will have an infinite number of additional solutions that are congruent to x (mod m), so we conventionally only bother to record the solutions x that lie in the range 0
