Finding X for Mod? If I have this:


*

*$x \pmod p = 1$

*$x \pmod q = 0$
Is there any way I can find a possible natural number for $x$ that satisfies both equations. I know it has something to do with the Chinese Reminder Theorem; however, I have been unable to solve it.
 A: No need for the Chinese Remainder Theorem. 

I'll assume you intended to require $p,q$ to be relatively prime positive integers.

Then by Bezout's Theorem, there exist integers $a,b$ such that 
$$ap+bq=1$$
If $p,q$ are given, qualifying values of $a,b$ can be found via the Extended Euclidean Algorithm.

Moreover, if $(a,b)$ is any qualifying pair, so is $(a',b')=(a-qt,b+pt)$, for any integer $t$.

Hence, by choosing $t$ appropriately, we can force $b' > 0$. 

Thus, we can get
$$a'p+b'q=1$$
with $b' > 0$.

Then letting $x=b'q$, it follows that $x$ is a positive integer such that
\begin{align*}
x&\equiv 1\;(\text{mod}\;p)\\[4pt]
x&\equiv 0\;(\text{mod}\;q)\\[4pt]
\end{align*}
A: $x\equiv 0\pmod{\!q}\!\iff\! q\mid x\!\iff\! x = qn,\ $ so $\bmod p\!:\ qn = x\equiv 1\!\iff\!\!\!\!\!\!\!\!\!\overbrace{ n\equiv q^{-1}}^{\large{\rm exists} \iff (q,p)=1}$
If so $\, x \equiv q(q^{-1}\bmod p)\, \pmod{\! pq}.\,$  Inverses van be computed via Extended Euclidean algorithm
It's a special case of CRT = Chinese Remainder Theorem (but we don't need that generality here)
A: Hint:
If you mod something by P and Q, what kind of number gives a result as a 1 or 0? think of Odd's and Even's.
