Solve the linear congurence $17x\equiv 3(\mod{2*3*5*7})$ by solving the system:
$17x\equiv 3(\mod{2})$
- For this one, I simplified to $x\equiv 1(\mod{2})$. Let this $x=5$.
$17x\equiv 3(\mod{3})$
- For this one, I simplified to $x\equiv 0(\mod{3})$. Let this $x=3$.
$17x\equiv 3(\mod{5})$
- For this one, I simplified to $2x\equiv 3(\mod{5})$, then again to $2x\equiv -2(\mod{5})$ and once more to $x\equiv -1(\mod{5})$.Let this $x=9$.
$17x\equiv 3(\mod{7})$
- For this one, I simplified to $3x\equiv 3(\mod{7})$, then again to $x\equiv 1(\mod{7})$. Let this $x=8$.
After all of this, I plugged it into the equation $$x\equiv a_1N_1x_1 + a_2N_2x_2 + a_3N_3x_3 + a_4N_4x_4 (\mod 210)$$
In the equation above, $N_1=105$, $N_2=70$, $N_3=42$, $N_4=30$. Also, $a_1=1$, $a_2=0$, $a_3=-1$ and $a_4=1$.
And I'm coming to $x\equiv177(\mod 210)$, and I should be coming to $x\equiv99(\mod 210)$. Can anyone tell me where I went wrong? It's driving me crazy.
Thanks for any help!
http://mathworld.wolfram.com/ChineseRemainderTheorem.html
has the notation I am about to use.
As you have computed correctly, $a_i ={1,0,-1,1}$ and $M/m_i= {105,70,42,30}$ (in order), now you must compute $b_i$ as the page describes. These are $b_i =1,1,3,4$ yielding $x \equiv 105*1*1 +0 -1*3*42 +1*30*4 = 99$ as desired.