# Thread: equation mod 195

1. ## equation mod 195

What's the quickest way to solve x^3 = 38 (mod 195)?

I see that 195 = 3*5*13
I think I should use the Chinese Remainder Theorem.
I find that
in mod 3, the solution is 2,
in mod 5, the solution is 2,
in mod 13, the solutions are 4,10,12.
So there should be 3 solutions to the given problem. However, I don't know how to proceed from here. The cubic part and the fact that I can break it up into 3 primes is a little different than the other ones in the text.

Thanks for any help.

2. Originally Posted by PvtBillPilgrim
What's the quickest way to solve x^3 = 38 (mod 195)?

I see that 195 = 3*5*13
I think I should use the Chinese Remainder Theorem.
I find that
in mod 3, the solution is 2,
in mod 5, the solution is 2,
in mod 13, the solutions are 4,10,12.
So there should be 3 solutions to the given problem. However, I don't know how to proceed from here. The cubic part and the fact that I can break it up into 3 primes is a little different than the other ones in the text.
Just treat it as three separate problems. You can combine the first two congruences to say $x\equiv2\!\!\!\pmod{15}$. Now use the CRT to find a solution mod 195 for each of these three problems:

(i) $x\equiv2\!\!\!\pmod{15},\quad x\equiv4\!\!\!\pmod{13}$,

(ii) $x\equiv2\!\!\!\pmod{15},\quad x\equiv10\!\!\!\pmod{13}$,

(iii) $x\equiv2\!\!\!\pmod{15},\quad x\equiv12\!\!\!\pmod{13}$.