Question

Find the remainder when ${{6}^{461}}$is divided by $7$ _________.
$A)1$
$B)5$
$C)6$
$D)4$

Answer 1

Hint: To solve this question we should have a knowledge of Binomial Expansion Theorem. The Theorem states that the expansion of any power having two numbers in addition or subtraction ${{\left( x+y \right)}^{n}}$ of a binomial $\left( x+y \right)$ as a certain sum of products. We will also be required to see the power of the number given to us. On substituting the number on the formula we will find the remainder.

Complete step-by-step solution:
The question asks us to find the remainder when a number which is given in the problem which is ${{\left( 6 \right)}^{461}}$, is divided by $7$. The first step is to write $6$ as a difference or sum of two numbers. The number when written in binomial form should be such that one of the numbers is divisible by $7$. On seeing the power of $6$, which is given as $461$, is an odd number. So the formula used will be:
$\Rightarrow {{\left( x-1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{\left( -1 \right)}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{\left( -1 \right)}^{1}}+.........+{}^{n}{{C}_{n}}{{x}^{0}}{{\left( -1 \right)}^{n}}$
Since the number $7$ is divisible by $7$, we will write $6$ as the difference between $7$and $1$. On substituting the number $7$ in place of $x$, and $461$ in place $n$ we get:
$\Rightarrow {{\left( 7-1 \right)}^{461}}={}^{461}{{C}_{0}}{{7}^{461}}{{\left( -1 \right)}^{0}}+{}^{461}{{C}_{1}}{{\left( 7 \right)}^{460}}{{\left( -1 \right)}^{1}}+.........+{}^{461}{{C}_{461}}{{\left( 7 \right)}^{0}}{{\left( -1 \right)}^{461}}$
On analysing the expansion we see that the value from the second term contains $7$ as one of their term, so the terms from second place will be divisible by $7$ as the number $7$ is divisible by $7$, so the number which is not divisible by $7$ is just the first term. The expansion gives us:
 $\Rightarrow {}^{461}{{C}_{461}}{{7}^{0}}{{\left( -1 \right)}^{461}}=-1$
Now writing it in terms of $7$ we get:
$\Rightarrow 7k-1$
The above expression could be further written as:
$\Rightarrow 7k-1+6-6$
$\Rightarrow 7k-7+6$
$\Rightarrow 7\left( k-1 \right)+6$
$\Rightarrow 7\left( k-1 \right)+6$
On analysing the above expression we get $6$ as the remainder.
$\therefore $ Remainder of ${{6}^{461}}$ when divided by $7$ is Option $C)6$ .

Note: If we are asked to find the remainder of the number having even power then the same process will be applied for solving. But the formula will change to ${{\left( x+1 \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}{{\left( 1 \right)}^{0}}+{}^{n}{{C}_{1}}{{x}^{n-1}}{{\left( 1 \right)}^{1}}+.........+{}^{n}{{C}_{n}}{{x}^{0}}{{\left( 1 \right)}^{n}}$, and then the problem will be solved accordingly.