practice problems pending concept class

Q1. Is there a solution to 5x = 3 mod 11?
Yes.
Why?
Since 5,11 are co-prime we can always find 'x' s.t. 5.x = r mod 11 for all 'r' from 0 to 10.

Q2. Simple way to solve
x = 1 mod 3
x = 2 mod 5
x = 3k + 1
3k + 1 = 2 mod 5
3k = 1 mod 5
=> k = 2 mod 5
=> k = 5t + 2
=> x = 3(5t + 2) + 1 = 15t + 7 = 7 mod 15

Similarly
x = 0 mod 8
x = 59 mod 125
125p + 59 = 0 mod 8
5p + 3 = 0 mod 8
5p = 5 mod 8
p = 1 mod 8
So x = 125*1 + 59 = 184

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