Question

Using binomial theorem, prove that $6^n-5n$ always leaves remainder $1$ when divided by $25$.

Answer 1

Assume $6^n=(1+5{)}^n$

we know that$(a+b{)}^n{=}^n{C}_0{a}^n{+}^n{C}_1{a}^{n-1}b+----{+}^n{C}_{n-1}a{b}^{n-1}{+}^n{C}_n{b}^n$

$\Rightarrow (1+5{)}^n{=}^n{C}_0(1{)}^n{+}^n{C}_1\left(1{)}^{n-1}\right(5\left){+}^n{C}_2\right(1{)}^{n-2}(5{)}^2+----{+}^n{C}_n(5{)}^n$

$\Rightarrow 6^n=1{+}^n{C}_1\left(5{)}^1{+}^n{C}_2\right(5{)}^2{+}^n{C}_3(5{)}^3+----{+}^n{C}_{n-1}(5{)}^{n-1}{+}^n{C}_n(5{)}^n$

$\Rightarrow 6^n=1+n\left(5\right){+}^n{C}_2\left(5{)}^2{+}^n{C}_3\right(5{)}^3+----{+}^n{C}_{n-1}\left(5{)}^{n-1}{+}^n{C}_n\right(5{)}^n$

$\Rightarrow 6^n=1+5n+\left(5{)}^2{[}^n{C}_2{+}^n{C}_3\right(5\left){+}^n{C}_4\right(5{)}^2+----{+}^n{C}_n\left(5{)}^{n-2}\right]$

$\Rightarrow 6^n=1+5n+25\left(k\right)$

where $k{=}^n{C}_2{+}^n{C}_3\left(5\right)+---{+}^n{C}_n(5{)}^{n-2}$

$\Rightarrow 6^n-5n=1+25k$ Hence $6^n-5n$ always leave remainder $1$ when dividing by $25$ Hence proved