Solving a Linear Congruence I've been trying to solve the following linear congruence with not much success:
$19\equiv 21x\pmod{26}$
If anyone could point me to the solution I'd be grateful, thanks in advance
 A: HINT: $21\equiv-5\pmod{26}$, so $21\cdot5\equiv-25\equiv1\pmod{26}$, and $21\cdot5\cdot19\equiv1\cdot19\pmod{26}$.
A: If $19 \equiv 21 x \mod 26$ then we must have $19=21x+26y$ for some integer $y$. You could use the Euclidean Algorithm to solve this problem, finding integers $s,t$ such that $\gcd(21,26) = 21s+26t$
A: $21^{-1}\equiv5\mod26\implies x\equiv95\equiv17\mod26$
A: Hint: $26 = 2 \cdot 13$ and the Chinese remainder theorem. Modulo $2$ we have to solve $1 \cong x \pmod 2$, that is $x = 2k + 1$ for some $k$, now solve $19 \cong 42k + 21 \pmod{13}$.
A: $$\frac{26}{21}=1+\frac5{21}=1+\frac1{\frac{21}5}=1+\frac1{4+\frac15}$$
The previous convergent of $\frac{26}{21}$ is $1+\frac14=\frac54$
Using Convergent property (Theorem #$3$ here) of continued fraction, $21\cdot5-26\cdot4=1\implies 21^{-1}\equiv 5\pmod {26} $
A: Hint: $\,{\rm mod}\ 26\!:\ \dfrac{19}{21} \equiv \dfrac{-7\cdot 1\ \ \,}{7\cdot 3}\equiv -\dfrac{27}{3}\equiv -9.\ \ \ $  Alternatively
note: $\,{\rm mod}\ 26\!:\ \dfrac{19}{21} \equiv \dfrac{-7}{-5} \equiv -7\left(\dfrac{-1}5\right) \equiv -7\left(\dfrac{25}5\right) \equiv -35 \equiv -9$
While such ad-hoc techniques often work quickly for small numbers, for larger numbers one should resort to algorithms such as the extended Euclidean algorithm.  Also, keep in mind that modular fractions uniquely exist only when the denominator is coprime to the modulus.
