What's the difference between these two solutions? Are both correct (task about modulus, extended euclidean algorithm)?

Find all solutions. You know that the multiplicative inverse is $2$.

$$17x \equiv 25 (\text{mod } 33)$$

First way of solving it:

Multiply this with $2$: $$34x \equiv 50(\text{mod } 33)$$

This is equivalent to:

$$1x \equiv 17 (\text{mod }33)$$

So solution is $x = 33k+17$ where $k \in \mathbb{Z}$

Second way of solving it:

Multiply this with $2$: $$34x \equiv 50(\text{mod } 33)$$

$$50 \text { mod } 33= 17$$

Thus $x = 17$

My question, are both solutions correct? And if yes, which one would you recommend? The second seems more comfortable for sure.

• The solutions are different, so how can they possibly both be correct? In particular, $x=17$ is just one solution, whereas you were asked to find all solutions. If there's more than one solution, your second answer must be wrong. – Gerry Myerson Feb 16 '17 at 11:05
• @GerryMyerson Thank you now I understood, totally forgot the word "all" mentioned here. – cnmesr Feb 16 '17 at 11:08

The suggested solution might have been $$17x\equiv 25\pmod{33}$$

$$\iff 50x\equiv 25\pmod{33}$$

$$\stackrel{:25}\iff 2x\equiv 1\pmod{33}$$

(which we can do because $\gcd(25,33)=1$)

$$\iff x\equiv 2^{-1}\pmod{33}$$

$$17x \equiv 25 \pmod{33}$$

The "natural" thing to do with $17x \equiv 25 \pmod{33}$ is to multiply both sides by what is $\dfrac{1}{17} \pmod{33}$. Since $2\cdot17 \equiv 34 \equiv 1 \pmod{33}$, then $\dfrac{1}{17} \equiv 2 \pmod{33}$. Multiply both sides by $2$ and you get

\begin{align} 34x &\equiv 50 \pmod{33} \\ x &\equiv 17 \pmod{33} \\ x &= 17 + 33n \quad (n \in \mathbb Z) \end{align}