# find the measure of $a,b,c,d,e,f,g$ and $h$.

if $\frac{1}{ab}+\frac{c}{de}+\frac{f}{gh}=1$ and they are different numbers between one and ten then find the measure of $a,b,c,d,e,f,g$ and $h$.

My Attempt:I find these answers with guessing.I know there isn't any way to find the answers without guessing but I want an easy guess system that the answer can found under 5 minutes.These are my answers:

$a=3$,$b=6$,$c=5$,$d=9$

$e=8$,$f=7$,$g=2$,$h=4$

update1:We cannot use 1 or 10.

update2:I want some one to explain the first part of @McFry solution.

• What do you mean by "measure"? – Anon Jun 18 '16 at 11:31
• the number that they are going to be. – Taha Akbari Jun 18 '16 at 11:32
• @TahaAkbari Please read this post meta.math.stackexchange.com/questions/3795/… It is bad form to delete and repost questions for the reasons explained in the link. – almagest Jun 18 '16 at 13:46

## 2 Answers

Consider what happens when you add some reduced fractions. For any prime power $r = p^a$, if exactly one of the denominators is divisible by $r$, then the denominator of the sum is also divisible by $r$. For $r=5$ and $r=7$ we see that we can't have $5$ or $7$ in the denominators, otherwise the sum wouldn't be 1.

Now consider $r = 9$. Since $9$ appears in one denominator and therefore divides it, another denominator must also be divisible by $9$. This is only possible when we have both $3$ and $6$ together in another denominator.

Now consider $r = 8$. Similarly, since $8$ appears in a denominator and $3\cdot 6$ is not divisible by $8$, we get that $2$ and $4$ must appear together in another denominator.

Putting this together, we get that the three numerators are $1, 5, 7$ and the three denominators are $3\cdot6,~ 2\cdot4$ and $8\cdot9$.

There are six ways to combine these things. It should be fairly easy to test all six ways in under 5 minutes.

• better answer +1.But maybe there is a way that we don't guess. – Taha Akbari Jun 18 '16 at 19:13
• Let's say $A$ is the numerator above $3\cdot 6$, $B$ is the numerator above $2\cdot 4$, and $C$ is the numerator above $8\cdot 9$. By multiplying the given equation by $72$, we get $4A+9B+C=72$. Take this equation mod $8$, and you get $4A+B+C\equiv 0$ mod $8$, and since $A$ is always odd, this simplifies to $B+C\equiv 4 \equiv 7+5$ mod $8$, so $A = 1$ is certain. Now modulo $9$, you get $0 \equiv 4A+C \equiv 4+C$, so $C$ must be $5$ and $B$ must be $7$. – Anon Jun 18 '16 at 19:57
• I mean why the other one should divisible by 9? – Taha Akbari Jun 19 '16 at 10:07
• That's explained in the first paragraph. Do you want me to elaborate? – Anon Jun 19 '16 at 10:38
• oh yes I know but please explain more that isn't the sum of them that is another. – Taha Akbari Jun 19 '16 at 10:53

We can restate the amended problem as follows. Divide the integers $1,2,3,4,5,6,7,8,9$ into three groups of three. From each group form a fraction by multiplying two of the numbers together to form the denominator and taking the third number as the numerator. 1 must be chosen as the numerator of its group. Set $S$ be the resulting set of three fractions. We have to show that only one such $S$ has sum 1. For example, you might take $S\{\frac{1}{2\cdot3},\frac{4}{5\cdot6},\frac{7}{8\cdot9}\}$, but that fails because its sum is not 1.

As @Arthur commented, 5 and 7 must appear as numerators, because none of the other numbers have a factor 5 or 7. So the three numerators are 1,5,7. We have to divide 2,3,4,6,8,9 into pairs to form the denominators. If we write the resulting fractions as $\frac{1}{A},\frac{5}{B},\frac{7}{C}$, then we require $ABC=BC+5AC+7AB$. Hence $A|BC,B|AC,C|AB$. That means we cannot pair 9 with 3 or 6, and we cannot pair 8 with 2,4 or 6.

So we must pair 8 with 9 or 3. If we pair 8 with 3, then we must pair 9 with 2 or 4. That gives two choices for the denominators: 24,18,24 or 24,36,12. The numerators are 1,5,7. Clearly 24,18,24 cannot work because the sum will be less than $\frac{1+5+7}{18}$. The largest possible sum in the case of 24,36,12 comes from matching 7 with the smallest denominator and 5 with the second smallest, but $\frac{7}{12}+\frac{5}{24}+\frac{1}{36}<1$, so there are no solutions from pairing 8 with 3.

So we must pair 8 with 9. That leaves three choices for the other denominators: 6,24 or 8,18 or 12,12. Again the numerators are 1,5,7. With 72,6,24 we cannot pair 7 with 6. So we have either $\frac{1}{6}+\frac{5}{24}+\frac{7}{72}=\frac{17}{36}$ or $\frac{1}{24}+\frac{5}{6}+\frac{7}{72}=\frac{35}{36}$ or $\frac{1}{6}+\frac{5}{72}+\frac{7}{24}=\frac{19}{36}$ or $\frac{1}{72}+\frac{5}{6}+\frac{7}{24}=\frac{41}{36}$.

With 72,8,18 if we pair 5 with 8, then the sum is either $\frac{1}{72}+\frac{7}{18}+\frac{5}{8}=\frac{37}{36}$ or much less if we pair 1 with 18. Clearly pairing 1 with 8 is even worse. So we must pair 7 with 8. Pairing 1 with 18 gives sum $\frac{1}{18}+\frac{5}{72}+\frac{7}{8}=1$. Pairing 1 with 72 will clearly give a different sum and fail.

Finally, the denominators could be 12,12,72 and the numerators 1,5,7. Taking 5 and 7 with 12 makes the sum too big, so that leaves two choices: $\frac{1+5}{12}+\frac{7}{72}=\frac{43}{72}$ or $\frac{1+7}{12}+\frac{5}{72}=\frac{53}{72}$. Thus we have the unique solution $$\frac{1}{3\cdot6}+\frac{5}{8\cdot9}+\frac{7}{2\cdot4}$$

That took me more than 20 minutes, concentrating fairly hard, to figure out and write up. I would never have attempted a similar exercise before the problem was simplified by the late update. The OP wanted a fast solution. In that case by far the quickest and most reliable approach is to code. It took me around 5 minutes to write up and run the code for the original problem. Adapted for the amended problem, it is as follows.

Note that I made no effort to use simplifications found by @Arthur or to discover others. It is faster just to run through all the possibilities. It still runs instantly through the 8! cases. The result apparently shows 16 cases, but they are all just rearrangements. The three fractions remain the same. However, the problem statement still (as I write this) fails to make clear that we are counting sets of fractions, rather than numbers of solutions in $a,b,c,d,e,f,g,h$.

Academic mathematicians tend to be dismissive of coding solutions, but there is always a trade-off between time and elegance. Of course, students need to learn how to do things by hand, so there is a good case for banning coding solutions for homework or exams.

• nice sloution +1. – Taha Akbari Jun 18 '16 at 15:46
• how did you know''$ABC=BC+5AC+7AB$. Hence $A|BC,B|AC,C|AB$'' – Taha Akbari Jun 18 '16 at 16:32
• You just multiply $1=\frac{1}{A}+\frac{5}{B}+\frac{7}{C}$ by $ABC$. – almagest Jun 18 '16 at 16:34
• I know how do you know after hence part. – Taha Akbari Jun 18 '16 at 16:35
• Sorry, each term in that equation has an $A$ in it except the term $BC$. If all the other terms are multiples of $A$, then $BC$ must be too. Similarly for the others. – almagest Jun 18 '16 at 16:36