The greatest integer $\mathrm{r}$ such that $30^{\mathrm{r}}$ divides 30 ! is
The greatest integer $\mathrm{r}$ such that $30^{\mathrm{r}}$ divides 30 ! is
$8$
$7$
$6$
$5$
Solution
Since $30^{\mathrm{r}}$ divides $30 !$
Now $30^{\mathrm{r}}=2^{\mathrm{r}} \times 3^{\mathrm{r}} \times 5^{\mathrm{r}}$ Since 5 divides 30 ! So $r=\left[\frac{30}{5}\right]+\left[\frac{30}{5^2}\right]=6+1=7$