Prove or disprove: there is an integer $x$ so that $x \equiv 2$ (mod 6) and $x \equiv 3$ (mod 9).

Question

Prove or disprove: there is an integer $x$ so that $x \equiv 2$ (mod 6) and $x \equiv 3$ (mod 9).

I'm not too sure how to approach this. I first noted that $(6,9) = 3 \neq 1$ so I cannot use the Chinese Remainder Theorem.

I also tried reaching a contradiction by supposing the above, and trying to show that one was odd and one was even, to no success.

I'm not sure how to proceed.

Answer 1

Hint: With the information, you have

• $3\mid 6$ and $6\mid x-2$.
• $3\mid 9$ and $9\mid x-3$.

Answer 2

$x \equiv 2\pmod 6 \implies x=6k+2, \ k\in \mathbb{Z}$

$x\equiv 3\pmod 9 \implies 6k+2 \equiv3\pmod9 \implies 6k\equiv1\pmod 9 \implies \color{blue}{\gcd(6,9)\mid 1}$

Answer 3

x≡2(mod 6) means that x has to be of the form x=6n+2(which is never divisible by 3)

and

x≡3(mod 9) means that x has to be of the form x=9m+3(which is always divisible by 3)

Answer 4

$$x\equiv2\pmod6\implies x\equiv2\pmod3$$

$$x\equiv3\pmod9\implies x\equiv3\pmod3\equiv0$$

But $2\not\equiv0\pmod3$