Filling in fractions I am trying to solve the problem in the below photo  

My try:
Let $S_N$ be the set $\{N,2N,3N\}$. These numbers will be the deominators of the fractions. $S_{11},S_{12},\cdots, S_{15}$ would not work as deominators as they use the same digit(s) twice or more. $S_{16}$ will not work either as the digits $5$ and $7$ are used as numerators. Both these digits need to be in the fraction with the smallest deominator or the fractions do not have the same deominators and the sum cannot be $1$. As for the others, when including the constraint that all of the deominaters can be simplifyed to $N$, none of the sets $S_{17},S_{18},S_{19}$ could make a sum larger than $1$. The sets $S_{20},S_{21},\cdots, S_{33}$ reuse a digit/digits, and $S_{34}$ and over contain three digit numbers.

By then, I have no more clues as to which numbers I should have as deominators.
Update: It turns out that up to the obvious permutations this problem has a unique solution whether you interpret the adjacent digits in the denominator as multiplication or concatenation. Who knew math could be so much fun.

Multiplication

I'm not sure yet of a better way to go about this, but for fractions to add up to an integer, you need the denominators to be "compatible" in some sense. E.g., you can't add some number of fifths with some number of sevenths and get an integer unless you started with two integers. With that in mind, your denominators should have most, if not all of the small primes ($2$ and $3$ for this problem).


  $$\frac1{3\cdot6}+\frac5{8\cdot9}+\frac7{2\cdot4}$$


Concatenation

The guiding principle is the same here. You need compatible denominators, so you want them to all have a roughly even distribution of the small primes in your problem. This is hampered partially by the difficulties in finding the prime factors of the sum of two numbers ($10a+b$ for digits $a$ and $b$), but the numbers in question are small and there aren't many of them, so just throwing everything even and most everything with a factor of $3$ into the denominator and hoping for the best reduces the number of things you have to check substantially.


  $$\frac5{34}+\frac7{68}+\frac9{12}$$
Update: It turns out that up to the obvious permutations this problem has a unique solution whether you interpret the adjacent digits in the denominator as multiplication or concatenation. Who knew math could be so much fun.

Multiplication

I'm not sure yet of a better way to go about this, but for fractions to add up to an integer, you need the denominators to be "compatible" in some sense. E.g., you can't add some number of fifths with some number of sevenths and get an integer unless you started with two integers. With that in mind, your denominators should have most, if not all of the small primes ($2$ and $3$ for this problem).


  $$\frac1{3\cdot6}+\frac5{8\cdot9}+\frac7{2\cdot4}$$


Concatenation

The guiding principle is the same here. You need compatible denominators, so you want them to all have a roughly even distribution of the small primes in your problem. This is hampered partially by the difficulties in finding the prime factors of the sum of two numbers ($10a+b$ for digits $a$ and $b$), but the numbers in question are small and there aren't many of them, so just throwing everything even and most everything with a factor of $3$ into the denominator and hoping for the best reduces the number of things you have to check substantially.


  $$\frac5{34}+\frac7{68}+\frac9{12}$$