Linear congruences $2X\equiv9\pmod{26},\pmod{25}$

Question

May double that of a natural number let rest $9$ when divided by $26$? And when divided by $25$?

I tried:

$$2X\equiv9\pmod{26}$$ As $(26,2)=2$ and $2\nmid9$ then the congruence linear not admits solution. So far, everything ok!

However in $$2X\equiv9\pmod{25}$$ As $(25,2)=1$ we assume that the congruence a single solution.

I tried to find the solution using Diophantine equations, however, observe $$2X\equiv9\pmod{25}\Longrightarrow25\mid2X-9\Longrightarrow\\2X-9=25k\Longrightarrow\fbox{$2X-25k=9$}$$

I'm on the right track?

How do I proceed if yes.

Answer 1

You're making it too complicated, I'm afraid. $2X\equiv9\pmod{25}$ is equivalent to $2X\equiv34\pmod{25}.$ Since $(2,25)=1,$ then we can conclude that $2$ has a multiplicative inverse modulo $25$ ($13$, in particular, but it isn't necessary to know this to complete the problem). This allows us to effectively "divide by $2$" on both sides of the congruence $2X\equiv34\pmod{25},$ giving us the answer.

Answer 2

Hint

$$2 \cdot 13 \equiv 26 \equiv 1 \pmod{25}$$

So, multiplication by 13 cancels $2$ in $\pmod{25}$.

Alternately Use the Extended Euclidian Algorithm to solve $2a+25b=1$ and multiply it by $9$ to get a solution to your Diophantine equation.