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

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?

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

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

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.

• thank you, but what is wrong with the way that I broke down the congruences? I used the theorem that I wrote in the post
– CnR
Jun 5, 2016 at 11:21
• Because $24=2^3\times3$, the Chinese remainder theorem allows you to check the congruence for the maximal prime powers dividing $24$, which are $2^3$ and $3^1$, i.e. $8$ and $3$. Not $4$ and $3$. Jun 5, 2016 at 11:24
• How do I decide to break down? I mean why couldn't it be p congruent to 2 mod 8 and p congruent to 3 mod 3?
– CnR
Jun 5, 2016 at 12:17
• Because $p\equiv11\pmod{24}$ implies that $p\equiv11\pmod{3}$ and $p\equiv11\pmod{8}$, by an argument similar to the one in my answer. Jun 5, 2016 at 13:46