1. congruences

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).

2. $2x \equiv 1 (mod7)\\
x\equiv 3(mod5)\\
x\equiv 3(mod8)$

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...)

3. Originally Posted by anncar
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).
See here.

First let's solve the first congruence for x:
$2x \equiv 1~\text{mod 7}$

$4 \cdot 2x \equiv 2 \cdot 1~\text{mod 7}$

$8x \equiv 2~\text{mod 7}$

$x = 2~\text{mod 7}$

So we need to solve
$x = 2~\text{mod 7}$
$x = 3~\text{mod 5}$
$x = 3~\text{mod 8}$

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
$5r_1 + (7 \cdot 8)s_1 = 1$

One such pair is $r_1 = -11$ and $s = 1$.

Now do the same for
$7r_2 + (5 \cdot 8)s_2 = 1$ <-- r2 = -17 and s2 = 3
and
$8r_3 + (5 \cdot 7)s_3 = 1$ <-- r3 = -13 and s3 = 3

Now, define $e_1 = (7 \cdot 8)s_1 = 56$, $e_2 = (5 \cdot 8)s_2 = 120$, and $e_3 = (5 \cdot 7)s_3 = 105$.

Then a solution to the system will be:
$x = 3 \cdot 56 + 2 \cdot 120 + 3 \cdot 105 = 723$
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 $723 \equiv 163~\text{mod 280}$. So we can use x = 163 as the solution.

-Dan

(See, you can teach an old dog new tricks!)

4. Oh, my apologies, I looked at it as three independent equations, just ignore my earlier post.