Question

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

Answer 1

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

We know that

$(a+b{)}^n{=}^n{C}_0{a}^n{b}^0{+}^n{C}_1{a}^{n-1}{b}^1+.....{+}^n{C}_n{a}^{n-n}{b}^n$

Putting $a=1,b=5$

$(6{)}^n{=}^n{C}_0{1}^n{5}^0{+}^n{C}_1{1}^{n-1}{5}^1{+}^n{C}_2{1}^{n-2}{5}^2+....{+}^n{C}_n{1}^{n-n}{5}^n$

${=}^n{C}_0{5}^0{+}^n{C}_1{5}^1{+}^n{C}_2{5}^2+....{+}^n{C}_n{5}^n$

$=1\times 1+\frac{n!}{1!(n-1)!}{5}^1+\frac{n!}{2!(n-2)!}{5}^2+....+1\times 5^n$

$=1+\frac{n(n-1)!}{1!(n-1)!}{5}^1+\frac{n(n-1)(n-2)!}{2!(n-2)!}{5}^2+....+1\times 5^n$

$=1+n\left(5\right)+\frac{n(n-1)}{2}{5}^2+....+5^n$

Thus, $(6{)}^n=1+5n+\frac{n(n-1)}{2}{5}^2+....+5^n$

$(6{)}^n-5n=1+\frac{n(n-1)}{2}{5}^2+....+5^n$

$(6{)}^n-5n=1+5^2(\frac{n(n-1)}{2}+....+5^{n-2})$

$(6{)}^n-5n=1+25(\frac{n(n-1)}{2}+....+5^{n-2})$

$(6{)}^n-5n=1+25k$

where $k=\frac{n(n-1)}{2}+....+5^{n-2}$

The above equation is of the form

Dividend $=$ Divisor $\times$ Quotient $+$ Remainder

$6^n-5n=25k+1$

Hence $6^n-5n$ always leave remainder 1 when dividing by 25.