If the fractional part of the number $$\frac{{{2^{403}}}}{{15}}$$  is $$\frac{k}{{15}},$$  then $$k$$ is equal to:

Solution :
$$\eqalign{ & {2^{403}} = {2^{400}} \cdot {2^3} \cr & = {2^{4\infty 100}} \cdot {2^3} \cr & = {\left( {{2^4}} \right)^{100}} \cdot 8 \cr & = 8{\left( {{2^4}} \right)^{100}} = 8{\left( {16} \right)^{100}} \cr & = 8{\left( {1 + 15} \right)^{100}} \cr & = 8 + 15\mu \cr} $$
When $${2^{403}}$$ is divided by 15, then remainder is 8. Hence, fractional part of the number is $$\frac{8}{{15}}$$ Therefore value of $$k$$ is 8