Find an improper fraction equal to $9$ Simple premise but I can't get anywhere with this problem!
The problem involves a fraction of the form $\frac{ABCDE}{FGHIJ}$ where $A,B,C,D,E,F,G,H,I,J$ are a number from $0-9$, each being used only once. This improper fraction must be equal to $9$ and I believe there are 3 possible answers.
I began working out the number of possible combinations, but there were thousands so I decided that was a dead end! Then I started looking into excel solutions but again I hit a wall. Any help would be appreciated.
Edit: I also believe that the $0$ is not allowed at the start of the 5 digit numbers i.e $A,F\neq 0$
 A: Here are some reductions that should hopefully be able to get you started. They are motivated by pretending the order of $A,B,C,D,E,F,G,H,I,J$ chosen is "random," and figuring out what sorts of things could stop the quotient from being $9$.

*

*You should expect the quotient to be roughly $1$, and, since it's the quotient of two $5$-digit numbers (two numbers of roughly the same size), it really shouldn't be that much larger than $1$. In particular, if $F\neq 1$, then $FGHIJ\geq 2\cdot 10^4$ and then $ABCDE \geq 18\cdot 10^4$, which can't occur. So, $F$ must be $1$, and $A$ must be $9$.


*In addition, for the numerator to be $<10^5$, we need the denominator to be $<\frac{10^5}{9}<12000$; since $G$ cannot also be $1$, $G$ must be $0$.


*$9$ must certainly divide the numerator. This means that $9|A+B+C+D+E$, and since $$9|A+B+C+D+E+F+G+H+I+J=45,$$ we also know that $9|F+G+H+I+J$, and thus the denominator is divisible by $9$ as well. So, the numerator is divisible by $81$. The fact that it lies between $9\cdot 10^4$ and $10^5$, has all distinct digits, and can't have a digit of $0$ or $1$ should narrow it down a fair bit.
A: Well I think you can guarantee that $A=9$ and $F=1$, just because otherwise would make it impossible to have it end up reducing to $9$ (it would be too small), unless you can say that $F=0$. For the rest, you can narrow it down and see that the numerator has to be AT LEAST divisible by $9$ (because it ends up reducing to $9$), but otherwise, I don't have anything for you.
