Solve the simultaneous congruences
2x = 1 (mod7)
x= 3(mod5)
x= 3(mod8)
those aren't supposed to be equal signs- they are three slashes(congruences).
$\displaystyle 2x \equiv 1 (mod7)\\
x\equiv 3(mod5)\\
x\equiv 3(mod8)$
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...)
See here.
First let's solve the first congruence for x:
$\displaystyle 2x \equiv 1~\text{mod 7}$
$\displaystyle 4 \cdot 2x \equiv 2 \cdot 1~\text{mod 7}$
$\displaystyle 8x \equiv 2~\text{mod 7}$
$\displaystyle x = 2~\text{mod 7}$
So we need to solve
$\displaystyle x = 2~\text{mod 7}$
$\displaystyle x = 3~\text{mod 5}$
$\displaystyle x = 3~\text{mod 8}$
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
$\displaystyle 5r_1 + (7 \cdot 8)s_1 = 1$
One such pair is $\displaystyle r_1 = -11$ and $\displaystyle s = 1$.
Now do the same for
$\displaystyle 7r_2 + (5 \cdot 8)s_2 = 1$ <-- r2 = -17 and s2 = 3
and
$\displaystyle 8r_3 + (5 \cdot 7)s_3 = 1$ <-- r3 = -13 and s3 = 3
Now, define $\displaystyle e_1 = (7 \cdot 8)s_1 = 56$, $\displaystyle e_2 = (5 \cdot 8)s_2 = 120$, and $\displaystyle e_3 = (5 \cdot 7)s_3 = 105$.
Then a solution to the system will be:
$\displaystyle x = 3 \cdot 56 + 2 \cdot 120 + 3 \cdot 105 = 723$
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 $\displaystyle 723 \equiv 163~\text{mod 280}$. So we can use x = 163 as the solution.
-Dan
(See, you can teach an old dog new tricks!)