<s>[INST] Integers and Fractions. Each of the numbers $2, 3, 4, 5, 6, 7, 8$ and $9$ is used once to fill in one of the
variables in the equation below to make it correct. Of the three fractions being added, what is the value of the largest one?
$\dfrac{1}{ab}+\dfrac{c}{de}+\dfrac{f}{gh}=1$
I have tried putting in different combinations of the numbers $2$ to $9$ in the blanks, but it's not leading me anywhere. Is there a smarter way to answer this question?
 [/INST] Since no other numbers share a common divisor with $5$ and $7$, they must be in the numerator. Therefore, let $c = 5$ and $f = 7$:
$$\frac{1}{ab} + \frac{5}{de} + \frac{7}{gh} = 1$$
Let us start with the fraction with the biggest numerator: $\displaystyle \frac{7}{gh}$. However, $ \displaystyle \frac{7}{2 \cdot 3}$ already is greater than $1$. 
After the next largest choice, $\displaystyle \frac{7}{2 \cdot 4}$, the numbers left over are $3, 6, 8, 9$. There are $\displaystyle {4 \choose 2} = 6$ possibilities, but since addition is communitative, there are only $3$ possibilities. These are:
$$\frac{1}{3 \cdot 6} + \frac{5}{8 \cdot 9} = \frac{1}{8}$$
$$\frac{1}{3 \cdot 8} + \frac{5}{6 \cdot 9} \ne \frac{1}{8}$$
$$\frac{1}{3 \cdot 9} + \frac{5}{6 \cdot 8} \ne \frac{1}{8}$$
Therefore, $$\frac{1}{3 \cdot 9} + \frac{5}{6 \cdot 8} + \frac{7}{2 \cdot 4} = 1, $$
and the value of the largest fraction is $\frac{7}{8}$.
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