The sets $\mathrm{A}=\left\{\mathrm{z}\mid {\mathrm{z}}^{18}=1\right\}\text{ and }\mathrm{B}=\left\{\mathrm{\omega }\mid {\mathrm{\omega }}^{48}=1\right\}$ consider the set $\mathrm{C}=\{\mathrm{z\omega }\mid \mathrm{z}\in \mathrm{A}\text{ and }\mathrm{\omega }\in \mathrm{B}\}$ If 'k' is number of distinct elements in C then sum of digits in 'k' is____

$\mathrm{z\omega }={\mathrm{e}}^{\mathrm{i}\frac{2{\mathrm{k}}_1\mathrm{\pi }}{18}}{\mathrm{e}}^{\mathrm{i}\frac{2{\mathrm{k}}_2\mathrm{\pi }}{48}}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{\pi }}{144}\left(8{\mathrm{k}}_1+3{\mathrm{k}}_2\right)}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{r\pi }}{144}}$
R can take at most 144 distinct values and all values are attained using the periodicity of 2$\mathrm{\pi }$ For example 3 x 47 + 8 = 149
$\frac{149\times 2\mathrm{\pi }}{144}-2\mathrm{\pi }=\frac{2\mathrm{\pi }\times 5}{144}$
Which is not attained for 8k₁ + 3k₂ between 0 to 143.