The remainder  when  $7^n-6n-50(n\in N)$  when divided by 36 is ___

$$\begin{gathered} \text{ Expanding }{7}^{\mathrm{n}}\text{ as }(1+6{)}^{\mathrm{n}}\text{ by binomial theorem we get } \\ \begin{array}{c}{=}^n{C}_0{6}^n{+}^n{C}_1{6}^{n-1}+\dots \dots \dots \dots \dots {\cdots }^n{C}_{n-2}{6}^2{+}^n{C}_{n-1}6{+}^n{C}_n{6}^0\\ =\left[{ }^n{C}_0{6}^n{+}^n{C}_1{6}^{n-1}+\dots \dots \dots \dots \cdots \cdots {\cdot }^n{C}_{n-2}{6}^2\right]+6n+1\end{array} \end{gathered}$$

$$\begin{gathered} =\left[36\lambda \right]+6n+1\text{ ( }36\lambda \text{ has all the numbers are multiple of }36\text{ as every } \\ \begin{array}{c}\Rightarrow 7^{\mathrm{n}}-6\mathrm{n}-50=\left[36\lambda \right]+6\mathrm{n}+1-6\mathrm{n}-50=\left[36\lambda \right]-49\\ =\left[36\lambda \right]-72+23\end{array} \end{gathered}$$