The number of integral terms in the expansion of $\displaystyle (5^ \frac{1}{2}+7^\frac {1}{6})^ {642}$ is?

Expansion of $\left(5^{1 / 2}+7^{1 / 6}\right)^{642}$

General term $=\mathrm{T}_{\mathrm{r}+1}={ }^{642} \mathrm{C}_{\mathrm{r}}\left(5^{1 / 2}\right)^{642-\mathrm{r}}\left(7^{1 / 6}\right)^{\mathrm{r}}$

$={ }^{642} \mathrm{C}_{\mathrm{r}} 5^{\frac{642-\mathrm{r}}{2}} 7^{\mathrm{r} / 6}$

For term to be integral $\frac{\mathrm{r}}{6}$ should be integer $\Rightarrow \mathrm{r}=\mathrm{o}, 6,12,18, \ldots 642$ at these values

$\frac{642-\mathrm{r}}{2}$ is also integral

$\Rightarrow$ Number of integral terms are $=108$