Find an improper fraction equal to $9$

Question

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 $A,F\neq 0$

Answer 1

• Since $F$ cannot be $0$, the denominator is at least $10,000$. Therefore, the numerator is at least $90,000$. Since the numerator is a $5$-digit number which is at least $90,000$, we know that $A=9$.

• Since the numerator is a $5$-digit number, it is less than $100,000$, dividing this by $9$ shows that the denominator is at most $11,111$. Therefore, $F=1$ and $G=0$ or $1$.

• Since numbers cannot be repeated, $G\not=1$ and so $G=0$.

At this point, we know the values of $A$, $F$, and $G$ and the remaining digits are $2,3,4,5,6,7,8$.

• The smallest that the denominator can be is $10,234$ as all other digits are taken and the largest the denominator can be is $10,876$ since the larger digits are taken. Therefore, the numerator varies between $92,106$ and $97,884$. In particular, $B$ is between $2$ and $7$ and $H$ is between $2$ and $8$.

Now, let's do a bit of a case-by-case analysis.

• If $H$ is $2$, then the largest that the denominator could be is $10,287$. In this case, the numerator is $92,583$. Therefore, $B$ is at most $2$, but we already know that $B$ is at least $2$, so $B=2$ in this case. This is not possible since $H$ was already assumed to be $2$.

Therefore, $H$ cannot be $2$ and the smallest that the denominator can be is $10,324$, so $H$ is between $3$ and $8$.

• If $H$ is $3$, then the largest that the denominator could be is $10,387$. In this case, the numerator is $93,483$. Therefore, $B$ is at most $3$, but since $H$ is $3$ and $B$ is at least $2$, it must be that $B=2$. Therefore, the largest that the numerator could be is $92,876$. Dividing this by $9$, we find that the denominator is at most $10,319$, which contradicts the conclusion that we have that the denominator is at least $10,324$.

Therefore, $H$ cannot be $3$, so $H$ is between $4$ and $8$ and the smallest that the denominator can be is $10,423$. Multiplying by $9$ gives us that the numerator is at least $93,807$, so $B$ is between $3$ and $7$.

• If $H=4$, then the largest that the denominator could be is $10,487$. Therefore the largest that the numerator could be is $94,383$. Hence, we find that $B$ is at most $4$, but we already know that $H=4$, so that is not possible. Since $B$ is at least $3$, it must be that $B=3$. The largest that the numerator could be when $B=3$ is $93,876$. So, by dividing by $9$, the largest that the denominator could be is $10,430$. By checking all the values between $10,423$ and $10,430$ for the denominator, we see that none of them satisfy the requirements.

Therefore, $H$ cannot be $4$ and the smallest the denominator can be is $10,523$. In this case, the numerator is at least $94,788$. Therefore, $H$ is between $5$ and $8$ and $B$ is between $4$ and $7$.

• If $H=5$, then the largest the denominator could be is $10,587$. Therefore, the largest the numerator could be is $95,283$. Therefore, $B$ is at most $5$, but since $5$ is already taken, it must be that $B=4$. The largest that the numerator could be is then $94,876$. This means that the denominator is at most $10,541$. Therefore, we know that the denominator is between $10,523$ and $10,541$, which leaves only a few cases to check.

At this point, it is helpful to note that the units digits of either number can't be $0$ or $5$ because then the units digit of the other number is also $0$ or $5$. In addition, the units digit of either number can't be $1$ or $9$ since those have already been used. By careful inspection, there are only ten denominators to check when $H=5$.

Let's assume that you've dealt with the case where $H=5$ and checked all ten denominators. Therefore, we're assuming that $H$ is at least $6$, so the smallest the denominator could be is $10,623$. Then the numerator is at least $95,607$, so $B$ is between $5$ and $7$.

Now, you can continue with this pattern on up, you would next assume that $H=6$ and see if you can determine $B$. Then, find the largest that the numerator can be and use that to restrict the denominator. You'll end up with a few more numbers to check for each value of $H$, but by looking closely, you can cut out a lot of options for the denominator and make each calculation reasonable (I'll leave the rest to you).

You can also use @Carl Schildkraut's answer to really cut down the things that you need to check, to just a few for each possibility for $H$.

Answer 2

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 $9$.

1. 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 $9$.

2. 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 $0$.

3. $9$ 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.

Answer 3

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.