The number  $a_n=6^n-5n$  for  $n=1,2,3,\mathrm{......}$ when divided by 25 leave the remainder:

$$\begin{gathered} a_n=6^n-5n={(5+1)}^n-5n{=}^n{C}_0{5}^n{+}^n{C}_1{5}^{n-1}+\mathrm{....}{+}^n{C}_{n-1}\times 5+1-5n \\ =25{(}^n{C}_0{5}^{n-2}{+}^n{C}_1\times 5^3+\mathrm{...}{+}^n{C}_{n-2})+1 \\ \Rightarrow \mathrm{Remainder}=1. \end{gathered}$$