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.

You might want to check Euler's theorem:
$a^{\phi(n)} \equiv 1 \mod n$ for each $a$ which is coprime with n.
$\phi(25) = \phi(5^2) = 5^2-5=20$
So $7^{101} = 7^{100}*7 = (7^5)^{20}*7\equiv7 \mod 25$
In the same way $18^{101} \equiv 18 \mod 25$, so their sum is $17+8=25=0 \mod 25$, which means that the given expression is divisible by 25.
