Is the CRT applicable here?

Question

Find the x that satisfies the following system:

$$x = 3 \mod 2$$

$$x = 3 \mod 5$$

$$x = 3 \mod 7$$

CRT:

$$ M = 2 * 5 * 7 = 70 $$

$$ M1 = 5 * 7 = 35$$

$$ M2 = 2 * 7 = 14$$

$$ M3 = 2 * 5 = 10$$

$$N1 = 1$$

$$N2 = 4$$

$$N3 = 5$$

$$ 3 * 35 * 1 + 3 * 14 * 4 + 3 * 10 * 5 = 423$$

$$ 423 = 3 \mod 70$$

So $ x = 3 $ but this is not true because $3 = 1 \mod 2$ and not $3 = 3 \mod 2$

I am unable to see my mistake, could it be that CRT is not applicable here?

Answer 1

CRT is trivially applicable here. $3\equiv 1 \mod 2$ and $3\equiv 3 \mod 2$ are both true. You are confusing equivalence modulo $n$ with taking a residue modulo $n$.

As a side note, you don't need CRT for things like that. If you have equations $x\equiv n\mod m_i$ for some $m_i$, then trivially $x=n$ satisfies the congruences, even if the assumptions of the CRT are not satisfied.

Answer 2

$3 = 1 \mod 2$ is not true but $3 \equiv 1 \pmod {2}$ is true.

When talking about the modulate equation it should not be equal mark '=' but equivalence mark '$\equiv$'.

For second question: CRT is applicable here and $x\equiv3 \equiv 1 \pmod {70}$ is a solution for the system of equations.