Extending Euclid's Algorithm

$36x \equiv 18 \pmod{102}$

$\therefore 102 \mid 36x- 18$

$\therefore 6\times 17 \mid 6 \times 3 ( 2x - 1)$

$\therefore 17 \mid 3(2x-1)$

But $3$ and $17$ are coprime, so $17 \nmid 3$

$\implies 17 \mid 2x-1$

This gives $2x-1$ is divisible by $17$ . And minimum such positive integer is $17$ itself.

$2x-1 = 17 \implies x= \boxed{9}$

Alternative:

Take the smallest x such that $36x>102$

$36\times (3) \equiv 6 \pmod{102}$

Multiply both sides by three

$36\times (9) \equiv 18 \pmod{102}$

$36x \equiv 18\pmod{102}$

Can be expressed as,

$36x = 18 + 102k$ , where $k$ is an integer

Dividing both sides by $6$ , we get

$6x = 3 + 17k$

$6x \equiv 3\pmod{17}$

$2x \equiv 1\pmod{17}$

$2x \equiv 18\pmod{17}$

$x \equiv 9\pmod{17}$

Therefore the smallest is $\boxed9$

Trial and Error

$x$ has to be an integer.

$36x = 102+18 \Rightarrow \text{x is not an integer}$

$36x = 2 \times 102 + 18 \Rightarrow \text{x is not an integer}$

$36x = 3 \times 102 + 18 \Rightarrow 36x \space = \space 324 \Rightarrow \text{x is an integer}$

Hence, $x = 9$ .

36x=18+102t, t is an integer

2x=1+(17/3)t

2x-1=(17/3)t

-> t has to be minimum 3 for x to be an integer

-> 2x-1=17

x=9