mod cancellation: compute $\, n/2\bmod 6\, $ from $\,n \bmod m\,$ for even $n$ I am using the C++ language.
I want to calculate these 2 expressions:- 
In our case, $x= 100000000000000000$
Expression(1)
$$((3^x-1)/2)\mod7$$
The numerator  $3^x-1$ is always divisible by $2$(basic number theory)
I calculated the above expression using Extended-Euclid-Gcd algorithm.
The above algorithm only works when gcd(denominator,mod-value)=1...in our case $\gcd(2,7)=1$ . So we were able to calculate it using the above algorithm.
Expression(2)
$$((3^x-1)/2)\mod6$$
The numerator  $3^x-1$ is always divisible by $2$(basic number theory-again)
Now, how do I calculate the above expression as $\gcd(2,6)=2$ which is not equal to $1$  ?
 A: For your first case, the extended Euclidean algorithm is probably the best way in general, as long as the denominator and modulus are coprime.
For your second case, consider $4\div 2$ modulo $6$. What should that be? Should it be $2$, as $2\cdot 2\equiv 4$? Or should it perhaps be $5$, as $5\cdot2\equiv 4$? There just isn't a single answer, and thus you can't really divide by $2$ modulo $6$.
If you're interested in what $\frac k2$ corresponds to modulo $6$, you have to find $k$ modulo $2\cdot 6=12$ and work your way from there. For instance, if $k\equiv_{12}4$, then $\frac k2$ is either $2$ or $8$ modulo $12$, meaning it must be $2$ modulo $6$.
A: Dividing by $\boldsymbol{q}$ when the modulus is divisible by $\boldsymbol{q}$
Note that
$$
aq\equiv bq\pmod{pq}
$$
precisely when
$$
a\equiv b\pmod{p}
$$
Thus, for $n\gt0$,
$$
3^n-1\equiv\left\{\begin{array}{}2&\text{if $n$ is odd}\\8&\text{if $n$ is even}\end{array}\right.\pmod{12}
$$
implies
$$
\frac{3^n-1}2\equiv\left\{\begin{array}{}1&\text{if $n$ is odd}\\4&\text{if $n$ is even}\end{array}\right.\pmod{6}
$$

A More Detailed Answer
This can be solved mod $6$ by solving mod $3$ and mod $2$, then applying the Chinese Remainder Theorem.
For $n\gt0$, $3^n\equiv0\pmod3$; therefore,
$$
\frac{3^n-1}2\equiv1\pmod3
$$
To compute
$$
\frac{3^n-1}2\pmod2
$$
we can to look at
$$
3^n-1\equiv(-1)^n-1\pmod4
$$
to get
$$
\frac{3^n-1}2\equiv\frac{(-1)^n-1}2\equiv\left\{\begin{array}{}0&\text{if $n$ is even}\\1&\text{if $n$ is odd}\end{array}\right.\pmod2
$$
Then the Chinese Remainder Theorem says we can combine the results mod $3$ and mod $2$ to get
$$
\frac{3^n-1}2\equiv\left\{\begin{array}{}4&\text{if $n$ is even}\\1&\text{if $n$ is odd}\end{array}\right.\pmod6
$$
A: Set 
$$y = \frac{3^x - 1}{2}.$$
Let $0 \leq b < 6$ such that $y = 6a + b$ for some integer $a$. Then
$$2y = 12a + 2b$$
and $0 \leq 2b < 12$, so you could simply compute $3^x - 1 \mod{12}$.
A: For even $\,n,\,$ if we wish to compute $\ n/2\bmod 6\ $ from $\ n\bmod m\ $ then we must have $\,12\mid m,\,$ i.e. we need to double the modulus to balance the division by $\,2,\,$ namely
$$\begin{align} {\rm notice}\ \ \ \,  \color{}{n/2 \equiv r\!\!\!\pmod{\!6}}&\!\iff n\equiv 2r\!\!\!\!\pmod{\!12}\\[.4em]
{\rm because}\ \ \  n/2\ =\ r\ +\ 6\,k\ \  &\!\iff n = 2r\ +\ 12\,k\end{align}\qquad$$
So first we compute $\,2r = n\bmod 12,\, $ therefore $\ r = n/2 \bmod 6\ $ by above, i.e.
$$\begin{align} r\,=\, \color{#c00}{n/2\bmod 6} \,&=\  \color{#c00}{(n\bmod 12)/2}\\[.4em] {\rm e.g.}\ \ \ 18/2 \bmod 6\, &= (18 \bmod 12)/2 = 6/2 = 3\end{align}\qquad$$
Beware $ $ We get the wrong result $\,0\,$ if we use $18\bmod 6$ vs. $18\bmod 12.\,$ See here for more.
Remark $ $ Below is a simpler way to compute $\,3^x\bmod 12\ $ (vs. using CRT mod $3\ \& 4)$ 
$$ 3^{K+{\large 1}}\!\bmod 12 = 3(3^K\!\bmod 4) = 3((-1)^K\!\bmod 4) = 3\ \ {\rm if}\ \ 2\mid K, \ {\rm else}\, \ 9$$
where we used $\ ab\bmod ac = a(b\bmod c) = $ mod Distributive Law
Note that our formula above is a special case of this law, namely
$$ 2\mid n\,\Rightarrow\ \color{#c00}{n\bmod 12\, =\, 2(n/2\bmod 6)}\qquad\qquad\qquad $$
