Proving $x$ is divisible by $20$ I need to prove that $x$ divisible by $20$ if and only if $x=0\pmod4$ and $x=0 \pmod 5$
proving that if $x=0 \pmod 4$ and $x=0 \pmod 5$ than $x$ divisible by $20$ is by the Chinese theorem (am I right??)
But the other way - I dont understand why its not enough to be divisible only by one of them ($5$ OR $4$)?
P.S. How does it help me prove $(7^n+4*2^n+8^n-3^n)|20$?? How do I open this?

$20 = 4*5$ so if $x$ is divisible by $20$ then it is divisible by $4$ and $5$. And hence by definition $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$ 
Conversely, if $x \equiv 0 \pmod 4$ and $x \equiv 0 \pmod 5$  then $x = 4n$ and $x= 5m$ for some integers $n$ and $m$.
So $4n$ = $5m$. 
Now since $5$ does not divide $4$, it must divide $n$. Hence $n = 5l$ for some integer $l$.
Hence $x = 4n = 4*5*l$.
Hence $x$ is divisible by $20$.
