How to solve for multiple congruence's what aren't relatively prime. How could I go about solving 
$$x\equiv 1\mod 2$$
$$x\equiv 1\mod 3$$
$$x\equiv 1\mod 4$$
$$x\equiv 1\mod 5$$
$$x\equiv 1\mod 6$$
$$x\equiv 0\mod 7$$
I know that if I want to use the Chinese Remainder Theorem then I have to find a way for all the mod's to be relatively prime to each other but I am unsure which ones I can get rid of.  Any suggestions?
 A: $x\equiv1\pmod4$ implies $x\equiv1\pmod2$, while $x\equiv1\pmod6$ is equivalent to the two congruences $x\equiv1\pmod2$ and $x\equiv1\pmod3$. So you should be okay keeping only the congruences modulo $3$, $4$, $5$, and $7$.
A: In this case, the system is equivalent to the following:
$$
\begin{array}{rcl}
x & \equiv & 1 \qquad \mod \mathrm{lcm}(2,3,4,5,6) \\
x & \equiv & 0 \qquad \mod 7
\end{array}
$$
This way the Chinese remainder theorem is applicable.
A: As a general comment, @user62015, a system of congruences with moduli that are not relatively prime can still have solutions. Take
$$
\begin{cases}
x \equiv a \pmod{m}\\
x \equiv b \pmod{n}
\end{cases}
$$
with $m,n$ not both zero.
If this has a solution $x$, then there are $s, t$ such that $a + ms = x = b + n t$, or
$$
b-a = ms - nt,
$$
so for a solution to exist we need $(m, n) \mid b - a$. Conversely, if $(m, n) \mid b - a$, use Euclid to find $u, v$ such that
$$
(m,n) = m u - n v,
$$
and multiply by the integer $(b-a)/(m,n)$ to get
$$
b-a = m (u (b-a)/(m,n)) - n (v (b-a)/(m,n)),
$$
so that 
$$
x = a + m (u (b-a)/(m,n)) = b + n (v (b-a)/(m,n))
$$ 
is indeed a solution.
A: Hint $\rm\ 2,3,4,5,6\mid x-1\iff 60 = lcm(2,3,4,5,6)\mid x-1\ $ so, by Easy CRT, we have
$$\begin{array}{ll}\rm x\equiv 0\ \ (mod\ 7)\\ \rm x\equiv 1\ \ (mod\ 60)\end{array}\rm \!\iff x\equiv 1 + 60 \left[\frac{-1}{60}\ mod\ 7\right]\equiv -119,\ \ by\ \  mod\ 7\!:\, \frac{-1}{60}\equiv \frac{6}{-3}\equiv -2$$
Theorem (Easy CRT) $\rm\ \ $ If $\rm\ m,n\:$ are coprime integers then $\rm\ n^{-1}\ $ exists $\rm\ (mod\ m)\ \ $ and
$\rm\displaystyle\quad \begin{eqnarray}\rm x&\equiv&\rm\ a\ \ (mod\ m) \\
\rm x&\equiv&\rm\ b\ \ (mod\ n)\end{eqnarray}  \iff x\ \equiv\ b + n\ \bigg[\frac{a\!-\!b}{n}\ mod\ m\:\bigg]\ \ (mod\ mn)$
Proof $\rm\ (\Leftarrow)\ \ \ mod\ n\!:\,\ x\equiv b + n\ [\cdots]\equiv b\:,\ $ and $\rm\ mod\ m\!:\,\ x\equiv b + (a-b)\ n/n\: \equiv\: a\:.$
$\rm\ (\Rightarrow)\ \ $ The solution is unique $\rm\ (mod\ mn)\ $ since if $\rm\ x',x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ therefore $\rm\ m,n\ |\ x'-x\ \Rightarrow\ mn\ |\ x'-x\ \ $ since $\rm\ \:m,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = mn\:.\ \ $ QED
Remark $\ $ The optimization used above to combine the  moduli where $\rm\,x\,$ takes the same (constant) value $\rm\:x\equiv 1,\:$ is know as the constant case of CRT = Chinese Remainder theorem. It proves quite handy in practice. See here for more on this.
