problem proving this property of congruence and primes I've been working on this for a few days and I just can't seem to find a good proof for this.  
Given $a \equiv b\pmod{p_i}$, $i=1,2,3,\dots,n$ and $p_i$ is prime, show that $a \equiv b \pmod{p_1p_2p_3\dots p_n}$

We are told that $a-b$ is divisible by all of the $p_i$, where $i$ ranges from $1$ to $n$.  We want to show that $a-b$ is divisible by $p_1p_2\cdots p_n$.     
We must assume that the $p_i$ are distinct. If they are not, the result does not necessarily hold.  
So we want to show that for any $c$, if all the $p_i$ divide $c$, then $p_1p_2\cdots p_n$ divides $c$.
From the fact that $p_1$ divides $c$, we find that $c=p_1c_1$ for some $c_1$.
But $p_2$ divides $c$, so $p_2$ divides $p_1c$. But $p_2$ is a prime, and does not divide $p_1$, so $p_2$ divides $c_1$, say $c_1=p_2c_2$.
It follows that $c=p_1p_2c_2$.
But $p_3$ divides $c$. Since $p_3$ is prime, and $p_3$ does not divide $p_1$ and $p_2$, it follows that $p_3$ divides $c_2$. Thus $c_2=p_3c_3$ for some $c_3$.
It follows that $c=p_1p_2p_3 c_3$.
Continue. After a while, we find that $c=p_1p_2\cdots p_n c_n$, which is the required result.
Remark: You can make the argument a little nicer by doing a formal induction.
The result is trivially true for $n=1$. 
Suppose the result is true for $n=k$. We want to show that the result holds for $n=k+1$.
So we are told that $c$ is divisible by $p_1,p_2,\dots,p_k,p_{k+1}$. By the induction assumption, $c$ is divisible by $p_1p_2\cdots p_k$, so $c=(p_1p_2\cdots p_k)q$ for some $q$. From the fact that $p_{k+1}$ divides $(p_1p_2\cdots p_k)q$, and that $p_{k+1}$ does not divide any of the $p_i$, $i=1$ to $k$, we conclude that $p_{k+1}$ divides $q$. This essentially completes the proof.
