A question about the existence of Maybe a stupid question: 
Let $p_1,p_2,p_3$ be prime numbers and $0\leq x_1<p_2$, $0\leq x_2<p_1$,$0\leq x_3<p_3$. If $x_1$ is fixed, is there exist such $x_2,x_3$ satisfying that
$x_1p_1+x_2p_2+x_3p_1p_2=kp_3 \mod p_1p_2p_3$ for some $k$ that $0\leq k<p_1p_2p_3$ ($k$ can be any value as you choose in the domain $0\leq k<p_1p_2p_3$).  If so, i further need to know that, if $x_1$ is fixed, are $x_2,x_3,k$ fixed? Let me give a example: let $p_1=2,p_2=3,p_3=5$, then we set $x_1=1,x_2=1,x_3=0$, we have: $1\cdot2+1\cdot3+0\cdot6=5$. I wonder the above solution, i.e., $x_1=1,x_2=1,x_3=0$, when $x_1=1$ is fixed, is unique or not. 
If above holds, could anyone give me a simplified proof (or proof sketch)? Thanks very much!
 A: I assume the $p_i$ are meant to be distinct.  Let $y_i$ satisfy $$-x_1p_1 \equiv y_1 \mod p_1$$ $$-x_1p_1 \equiv y_2 \mod p_2$$ $$-x_1p_1 \equiv y_3 \mod p_3 \\$$Clearly $y_1 \equiv 0 \mod p_1$
Denote by $(a^{-1})_b$ the multiplicative inverse of $a \mod b$.  Then using the Chinese Remainder Theorem:
$-x_1p_1 \equiv \left[y_2p_1(p_1^{-1})_{p_2}(p_3^{-1})_{p_1p_2}\right]p_3+\left[y_3(p_1p_2)_{p_3}^{-1}\right]p_1p_2 \mod p_1p_2p_3$
Setting $k=-y_2p_1(p_1^{-1})_{p_2}(p_3^{-1})_{p_1p_2}$ and choosing any $x_2$ and $x_3$ satisfying $x_2'+x_3=y_3(p_1p_2)_{p_3}^{-1}$, where $x_2=x_2'p_1$, should solve it.  

Here's some more details on how I came up with that construction.  $$-x_1p_1 \equiv y_1 \mod p_1$$ $$-x_1p_1 \equiv y_2 \mod p_2$$
implies $$-x_1p_1 \equiv y_1p_2(p_2^{-1})_{p_1}+y_2p_1(p_1^{-1})_{p_2} \mod p_1p_2$$
by the Chinese Remainder Theorem.  Thus, $$-x_1p_1 \equiv y_2p_1(p_1^{-1})_{p_2} \mod p_1p_2$$ $$-x_1p_1 \equiv y_3 \mod p_3 $$
so again by the Chinese Remainder Theorem, 
$-x_1p_1 \equiv \left[y_2p_1(p_1^{-1})_{p_2}\right]p_3(p_3^{-1})_{p_1p_2}+\left[y_3\right](p_1p_2)(p_1p_2)_{p_3}^{-1} \mod p_1p_2p_3 \Rightarrow $ 
$-x_1p_1 \equiv \left[y_2p_1(p_1^{-1})_{p_2}(p_3^{-1})_{p_1p_2}\right]p_3+\left[y_3(p_1p_2)_{p_3}^{-1}\right]p_1p_2 \mod p_1p_2p_3 \Rightarrow$ 
$-x_1p_1 \equiv -kp_3+(x_2'+x_3)p_1p_2 \mod p_1p_2p_3 \Rightarrow$ 
$kp_3 \equiv x_1p_1+x_2'p_1p_2+x_3p_1p_2 \mod p_1p_2p_3 \Rightarrow$ 
$kp_3 \equiv x_1p_1+x_2p_2+x_3p_1p_2 \mod p_1p_2p_3$
