Does a solution to this congruent equation exist?

Question

Does a solution to this congruent equation exist? $$\begin{align} &\equiv 14 \pmod {30}\\ &\equiv 28 \pmod{66} \end{align}$$

Answer 1

Hint $\ 14+30j = x = 28+66k\,\Rightarrow\, 14 = 30j-66k = 3(\cdots)\,\Rightarrow\,3\mid 14$

i.e. $\ {\rm mod}\ 3\!:\ \color{#c00}{14\equiv x\equiv 28}\,\Rightarrow\, 3\mid 28\!-\!14\ $ since congruences persist mod factors of the modulus.

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Remark $\ $ Similarly if $\, x\equiv a\pmod m,\ x\equiv b\pmod n\,$ then $\, \color{#c00}{a\equiv x\equiv b}\pmod d\,$ for $\,d =\gcd(m,n),\ $ hence $\,d\mid a-b\ $ is a necessary condition for the existence of a solution.

This compatibility condition is also a sufficient condition for the existence of solution, and it extends pairwise to any number of congruences - see this answer for a constructive proof (which depends on the key fact that gcd distributes over lcm).

Answer 2

$$x\equiv14\pmod{30}\implies x\equiv14\pmod6\equiv2$$

$$x\equiv28\pmod{66}\implies x\equiv28\pmod6\equiv4$$

But $4\not\equiv2\pmod6$

Answer 3

There is no solution.

1) The two moduli $30, 66$ have greatest common divisor=$6$.

2) Take the required resudues modulo $6$, that number being the gcd given above. Thus in the first equation $x\equiv 14 \equiv 2$ mod 6$, in the second equation $x\eqiiv 28 \equiv 4$ mod 6$.

3) We can't have any solution congruent with both $2$ and $4$ modulo $6$. Bummer!