1. ## solve the system

x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)

2. hello,
Originally Posted by Sally_Math
x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
$\displaystyle 2x \equiv 3 \bmod 7 \equiv 10 \bmod 7 \Rightarrow x \equiv 5 \bmod 7$

$\displaystyle 3x \equiv 4 \bmod 11 \equiv 15 \bmod 11 \Rightarrow x \equiv 5 \bmod 11$

now use the chinese remainder theorem.

3. Originally Posted by Moo
hello,

$\displaystyle 2x \equiv 3 \bmod 7 \equiv 10 \bmod 7 \Rightarrow x \equiv 5 \bmod 7$

$\displaystyle 3x \equiv 4 \bmod 11 \equiv 15 \bmod 11 \Rightarrow x \equiv 5 \bmod 11$

now use the chinese remainder theorem.
ok i know that u start this way:-
using a=b+km (for eq. 1)
which is x=2+5k
then substituting into eq. 2, u get
2(2+5k)=3 (mod 7)
then we get
4+20k=3 (mod 7)

and then that's where i get stuck

4. Originally Posted by Sally_Math
x=2 (mod 5), 2x=3 (mod 7), 3x=4 (mod 11)
The congruence, $\displaystyle 2x\equiv 3(\bmod 7)$ is equivalent to $\displaystyle 4(2x)\equiv 4\cdot 3 (\bmod 7)$ which becomes $\displaystyle x\equiv 5(\bmod 7)$. The congruence, $\displaystyle 3x\equiv 4(\bmod 11)$ is equivalent to $\displaystyle 4(3x)\equiv 4\cdot 4(\bmod 11)$ which becomes $\displaystyle x\equiv 5(\bmod 11)$.

Thus, you have to solve,
$\displaystyle x\equiv 2(\bmod 5)$
$\displaystyle x\equiv 5(\bmod 7)$
$\displaystyle x\equiv 5(\bmod 11)$

Combine the last two congruences and so,
$\displaystyle x\equiv 2(\bmod 5)$
$\displaystyle x\equiv 5(\bmod 77)$

This is equivalent to,
$\displaystyle x\equiv 2 + 16\cdot 5(\bmod 5)$
$\displaystyle x\equiv 5 + 77(\bmod 77)$

We get,
$\displaystyle x\equiv 82(\bmod 5)$
$\displaystyle x\equiv 82(\bmod 77)$

Combine,
$\displaystyle x\equiv 82(\bmod 385)$