CRT general equal for three congruences - correct?

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I saw someone ask a question regarding the congruences below. $c≡1 \mod 143$

$c≡315 \mod 323$

$c≡167 \mod 667$

I tried to solve them with this equation: For any number of congruences $x≡a\mod n$, $N$ is the product of the moduli.

$c= \left[\Sigma_i = a_i \frac N{n_i} \left[\left(\frac N{n_i}\right)^{-1}\right]_{n_i}\right]_N$

I won't post all my the Extended Euclidean algorithm stuff but essentially these are the results I got:

$1\left( \frac {30808063} {143} \right)\left[\left( \frac {30808063} {143} \right)^{-1}\right]_{143}$ +

$315\left( \frac {30808063} {323} \right)\left[\left( \frac {30808063} {323} \right)^{-1}\right]_{323}$ +

$167\left( \frac {30808063} {667} \right)\left[\left( \frac {30808063} {667} \right)^{-1}\right]_{667}$

$ = (215441*122 + 315*95381*286 + 167*49189*221 )\mod 30808063 = 3154410$

Is this correct?