is $7^{101} + 18^{101}$ divisible by $25$? I am not able to find a solution for this question. I am thinking in the lines of taking out some common element like  $(7\cdot 7^{100}) + (18\cdot18^{100})$
but couldn't go anywhere further.

We can use lifting the exponent lemma. Let $v_5(n)$ be the exponent of the maximum power of $5$ dividing $n$.
$v_5(7^ {101}+18^{101})=v_5(25)+v_5(101)=2+0=2$.
So it is a multiple of $25$.
