If $$\frac{a}{b} = \frac{3}{4}, \frac{b}{c} = \frac{4}{5}  and  \frac{c}{d} = \frac{5}{6}$$, then the sum of the numerator and the denominator (which are coprimes) of $$\left(\frac{a}{d}\right)^{10}$$ is:

Given that $$\frac{a}{b} = \frac{3}{4}, \frac{b}{c} = \frac{4}{5} and \frac{c}{d} = \frac{5}{6}$$

Then, $$\frac{a}{d}=\frac{a}{b} \times \frac{b}{c} \times \frac{c}{d} $$

=> $$\frac{a}{d} = \frac{3}{4} \times \frac{4}{5} \times \frac{5}{6} $$

=>$$\frac{a}{d} = \frac{1}{2} $$

Therefore, sum of the numerator and the denominator (which are coprimes) of $$\left(\frac{a}{d}\right)^{10}$$ $$ = 1 + 2^{10} = 1025 $$