$(\sqrt{9})^{3} \times (\sqrt{81})^{5} \div 27^{2} = 3^{?}$

$A$ number exceeds its two-fifths by $75$. The number is:

$A$ number when divided by a certain divisor leaves a remainder of $241$. When twice the number is divided by the same divisor,the remainder is $112$. Find the divisor.

For any integral value of $n$,$3^{2n} + 9n + 5$ when divided by $3$ leaves the remainder:

$A$ number when divided by $221$ gives a remainder $43$. What remainder will be obtained by dividing the same number by $17$?

$A$ number consists of two digits such that the digit in the ten's place is less by $2$ than the digit in the unit's place. Three times the number added to $\frac{6}{7}$ times the number obtained by reversing the digits equals $108$. The sum of the digits in the number is: