Solutions for a system of congruence equations

Question

I have a system

$$ \begin{cases} x \equiv 7 \pmod{15} \\ x \equiv 14 \pmod{33} \end{cases} $$

How can I show that the system does not have any solutions?

I know that the first implies that $x = 7+15k$ and the second implies that $x = 14+33m$ and $\text{gcd}(15,33) = 3$.

Edit

Trying to show that another system has no solutions $$ \begin{cases} x \equiv 14 \pmod{98} \\ x \equiv 1 \pmod{28} \end{cases} $$

I have $\text{gcd}(98,28) = 14$.

From the system I get $x = 14+98k$ which implies that $x \pmod{14} = 0$. From the other congruence equation I get $x = 1+28m$ which implies that $x \pmod{14} = 1$. Therefore the system has no solutions.

But why am I using the trick by finding $x \pmod{\text{gcd}(n_1,n_2)}$ for each congruence equation? Is it just to find a number which can cancel out both $98$ and $28$?

Answer 1

Hint: what is the value of $x$ modulo $3$?

Answer 2

Consider

$x \equiv a (mod n)$
$ x \equiv a(mod m)$

Then the above has a solution only if gcd(m,n) divides gcd(a,b). Here gcd(15,33)=3 and gcd(7,14)=7. Here the given equation has no solution

Answer 3

For the first system:

"I know that the first implies that x=7+15k and the second implies that x=14+33m and gcd(15,33)=3."

We try to solve the system of equations! we get $7+15k=14+33m \rightarrow (15k-33m)=7 \rightarrow 3(5k-11m)=7$, which is impossible as 7 is not divisible by 3.

Additionally, $x=7+15k$ means that $x \cong 1 (mod 3)$. $x=14+33m$ means that $x \cong 2 (mod 3)$. Contradiction, hence no solutions.

The second system can be similarly solved by considering it mod 2.