Question

If $6^n-5n$ is divided by 25 , then remainder is

• A. 0
• B. 2
• C. 4
• D. 1

Answer 1

Answer: (D)

For two numbers a and $b$, if we can find numbers $q$ and $r$ such that $a=bq+r$, then we say that $b$ divides a with $q$ as quotient and $r$ as remainder. Thus, in order to show that $6^n-5n$ leaves remainder 1 when divided by 25 , we prove that $6^n-5n=25k+1$, where $k$ is some natural number.
We have,
$(1+a{)}^n={}^n{C}_0+{}^n{C}_{1}a+{}^n{C}_2{a}^2+\dots +{}^n{C}_n{a}^n$
For $a=5$, we get
$(1+5{)}^n={}^n{C}_0+{}^n{C}_{1}5+{}^n{C}_2{5}^2+\dots +{}^n{C}_n{5}^n$
i.e., $(6{)}^n=1+5n+5^2\cdot {}^n{C}_2+5^3\cdot {}^n{C}_3+\dots +5^n$
i.e., $6^n-5n=1+5^2\left({}^n{C}_2+{}^n{C}_{3}5+\dots +5^{n-2}\right)$
or $6^n-5n=1+25\left({}^n{C}_2+5\cdot {}^n{C}_3+\dots +5^{n-2}\right)$
or $6^n-5n=25k+1$, where $k={}^n{C}_2+5\cdot {}^n{C}_3+\dots +5^{n-2}$
This shows that when divided by $25,6^n-5n$ leaves remainder 1.