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.

Since 7 and 18 are coprime to 25, and the number of elements of the group of units of the ring Z/25Z is 20, 7^101 is congruent to 7 and 18^101 is congruent to 18 modulo 25 by Lagrange's theorem. This is essentially using Euler's extension of Fermat's little theorem, in the terminology of modern algebra.
