Solving $39x\equiv75\pmod{102}$

Question

Decide if a solution to the congruence $39x\equiv75\pmod{102}$ exists.

As $\operatorname{hcf}(39,102)=3$ we can write the congruence as $13x\equiv25\pmod{34}$. Using Euclid's algorithm gives $1=5\cdot34-13\cdot13$.

I don't get where to go from here.

Answer 1

As you wrote, $39x\equiv75\mod102\iff13x\equiv25\mod34.$

You also wrote $1=5\times34-13\times13$; that implies $1\equiv-13\times13\equiv21\times13\mod34$.

Therefore $x\equiv21\times25\equiv15\mod34$, so solutions are $x\equiv 15, 49, $ or $83\mod 102$.

Answer 2

Cancelling $3$ yields $\bmod 34\!:\,\ 13x\equiv 25\iff x\equiv \dfrac{25}{13}\equiv \dfrac{75}{39}\equiv \dfrac{75}5\equiv 15$ by Gauss's algorithm

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If you only need to prove that a solution exists then:

$$\begin{align} \exists x\!:\,\ &39x\equiv 75\!\!\!\pmod{\!102} \\ \iff \exists x,y\!:\,\ &39x - 102y=75\\ \iff\qquad\quad\ &(39,102)\mid 75,\ \ {\rm by\ \ Bezout}\\ \iff\qquad\quad\ &\qquad\quad\ 3\mid 75,\ \ {\rm by}\ \ (39,102) = 3(13,34) = 3\\ \text{The calculation above goes further, viz.}\\[.5em] \exists x\!:\,\ &39x\equiv 75\!\!\!\pmod{\!102} \\ \iff \exists x,y\!:\,\ &39x - 102y=75\\ \iff \exists x,y\!:\,\ &13x \,-\, 34y=25,\ \ \text{by cancelling } 3\\ \iff\quad \exists x\!:\,\ & 13x\equiv25\!\!\!\pmod{\!34}\\ \end{align}\qquad\qquad\qquad\qquad\qquad$$

See here for more on this connection between Bezout and (modular) divisibility.

Answer 3

The Bézout's relation you obtain: $5\cdot 34-13\cdot 13=1$ says that inverse of $13\bmod 34 $ is $-13$, so $$13x\equiv 25\pmod{34}\iff x\equiv (-13)\cdot 25=-325\equiv 15\pmod{34}.$$

Answer 4

You can actually reduce even more if you know $13\equiv -21\pmod {34}$ and $25\equiv -9\pmod {34}$ then you get:

$$-7x\equiv-3\pmod{34}$$ on division by 3, then $$7\cdot 5\equiv 1\pmod {34}\implies -7\cdot 5\equiv -1\pmod{34}$$$$\implies -7\cdot 15\equiv -3\pmod{34}$$