Im trying to solve the following Simultaneous congruence.
$2x ≡ 3(mod\ 5) $
$3x ≡ 2(mod\ 4)$
$4x ≡ 3(mod\ 9) $
by Chinese remainder theorem
$x$ = $B_1c_1x_1 \ + \ B_2c_2x_2 \ + B_3c_3x_3 \ $
Where
$c_1 = 3$
$c_2 = 2 $
$c_3 = 3$
$B_1 = 2$x$3 $
$B_2 = 3$x$3 $
$B_3 = 3$x$2 $
$x_n=b_n(mod \ b_1)$
$x_1 = 1$
$x_2 = 1 $
$x_3 = no \ solution $
This is all i know and this is when $ x = c (mod \ n)$
but since there is a coefficient infront of x what should i change?
Thanks in advance.
Best Answer
All coefficients in front of $x$ are coprime to their respective moduli, which means we can multiply by the inverses of said coefficients. Take the first equation: $$2x\equiv3\bmod5$$ The multiplicative inverse of 2 modulo 5 is 3. Thus $$x\equiv2x\cdot3\equiv3\cdot3\equiv4\bmod5$$ Similarly $$x\equiv2\bmod4$$ $$x\equiv3\bmod9$$ Now the Chinese remainder theorem applies, and we get $x\equiv174\bmod180$.