Solve the congruence system: $p \equiv 11\pmod{24}$ and $ p\equiv 3 \pmod 4$

Question

I want to find the solutions of the congruences system: $p \equiv 11\pmod{24}$ and $ p\equiv 3 \pmod 4$.

I probably have some mistake in my solution, can you tell me where I'm wrong?

$ 4 $ and $24$ are non co-primes so I can't use Chinese theorem. So I used this:

and becuase $24=2^3 * 3 $ I found the solution for the equivalent system: $p \equiv 2\pmod3$ and $ p\equiv 3\pmod 4 $ and now I used the Chinese theorem and found that $ p\equiv 11\pmod{12} $.

But when trying to check my answer, I noticed that for exmaple $ 23\equiv 11 \pmod{12}$ but it's not true that $ 23\equiv 11\pmod{24}$ which is the original congruence, so what did I do wrong? And could anyone explain to me the method to solve congruences like this where the modulos are not coprime?

Answer 1

$$p\equiv11\pmod{24}\implies p\equiv11\pmod4\equiv3$$

So, it is sufficient to have $$p\equiv11\pmod{24}$$

Answer 2

If $p\equiv11\pmod{24}$ then $p\equiv3\pmod{4}$, because if $p=11+24k$ for some $k\in\Bbb{Z}$ then $$p=3+4(2+6k),$$ with $2+6k\in\Bbb{Z}$. So you only need to solve $p\equiv11\pmod{24}$.

The equivalent system, using the Chinese Remainder theorem, should be $$p\equiv2\pmod{3}\qquad\text{ and }\qquad p\equiv3\pmod{8},$$ the latter congruence should be modulo $8$. Breaking down congruences this way is a good approach to solving systems of modular equations, also when the moduli are not coprime.