No simultaneous solutions (Chinese Remainder Theorem)

Question

a. Show that $x\equiv2\pmod6$ and $x\equiv3\pmod4$ have no simultaneous solutions.

If $x\equiv2\pmod6$ then x is even but if $x\equiv3\pmod4$ then x is odd. This is a contradiction, so $x\equiv2\pmod6$ and $x\equiv3\pmod4$ have no simultaneous solutions.

b. Show that $x\equiv2\pmod6$ and $x\equiv0\pmod4$ have no simultaneous solutions.

If $x\equiv2\pmod6$ then x is even and if $x\equiv0\pmod4$ then x is a multiple of 4, so it is also even, so $x\equiv2\pmod6$ and $x\equiv0\pmod4$ have simultaneous solutions.

Answer 1

If $x \equiv 2\pmod{6}$, then is $x$ even or odd?

If $x \equiv 3\pmod{4}$, then is $x$ even or odd?

Do you see the contradiction?