Solve the simultaneous congruences
2x = 1 (mod7)
those aren't supposed to be equal signs- they are three slashes(congruences).
I would do the first one, but I'm out of time, gotta get ready for work, and wouldn't want to steer you wrong.
For the second one, you can see that 3=5*0+3, so when x = 3, it is congruent to 3(mod5). So find the next in the progression by just adding 5, so x=3+5n for any integer n (this creates the series ...,3,8,13...)
For the third one, you can see that 3=8*0+3, so when x=3 it is congruent to 3(mod8). So find the next progression by just adding 8, so x=3+8n for any integer n (this creates the series ...3,11,19...)
First let's solve the first congruence for x:
So we need to solve
The modulos are pairwise coprime, so we may state a solution exists by the Chinese Remainder Theorem.
Use the extended Euclidean algorithm to find integer r1 and s1 such that
One such pair is and .
Now do the same for
<-- r2 = -17 and s2 = 3
<-- r3 = -13 and s3 = 3
Now, define , , and .
Then a solution to the system will be:
which you may verify is a correct solution.
Edit: Whoops! I just noticed. This answer will be correct a modulo 5 x 7 x 8 = 280. So we have a smaller answer which is equivalent to . So we can use x = 163 as the solution.
(See, you can teach an old dog new tricks!)