What is the largest possible value for the sum of two fractions such that each of the four 1 digit prime numbers occurs as one of the numerators or denominators?
Hello, Godfather!
Here's one approach . . .What is the largest possible value for the sum of two fractions such that
each of {2, 3, 5, 7} occurs as one of the numerators or denominators?
The sum of two fractions is: .$\displaystyle S\:=\:\frac{a}{b} + \frac{c}{d} \:=\:\frac{ad + bc}{bd}$
For the largest value, we want the smallest denominator.
. . This occurs for: .$\displaystyle b = 2,\:d = 3$ .(or vice versa).
We have: .$\displaystyle S \:=\:\frac{3a + 2b}{6}$
Then we have two choices: .$\displaystyle \begin{array}{c}a = 5,\,b = 7 \\ a=7,\,b=5\end{array}$
If $\displaystyle a=5,\,b=5:\;S\:=\:\frac{3\!\cdot\!5+2\!\cdot\!7}{6 } \:=\:\frac{29}{6}$
If $\displaystyle a=7,\,b=5:\;S\:=\:\frac{3\!\cdot\!7 + 2\!\cdot\!5}{6}\:=\:\frac{31}{6}\quad\Leftarrow\:\ text{larger!} $
Therefore: .$\displaystyle \frac{7}{2} + \frac{5}{3}\:=\:\boxed{\frac{31}{6}}$ .is the largest sum.