I said let:
Since $R$ is divisible by $360$ then it is also divisible by $2$. So $F=0,2,8,k$ only if $k$ is even.
$R$ is divisible by $5$ so the last digit $F=0/5$. So $F=0$
Also $R$ is divisible by $4$ so $EF$= divisible by $4$. but $F=0$ so, $EF=20,80$
Since it is divisible by $8$ then the last three digits form a number divisible by $8$. So $DEF=320, 280$
Now $ABC$ is also divisible by $8$ so, if $DEF=320$ then $ABC=18K$ if $K=4$. If $DEF=280$ then $ABC=13K$ if $K=6$
but then I noticed Also $R$ is divisible by $3$ and $9$, so i need the sum of the number to be a multiple of $3$. I have so far
$R=18K320=184320$ where the sum of the digits is $18 $ which is a multiple of $3$ and $9$
I also have:
$R=13K280=136280$ where the sum of the digits is $20$ which is not a multiple of $9$ or $3$. So this number does not work.
So I need help finding two more numbers. Any ideas?