Answer

Verified

348.9k+ views

**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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Establish a relation between electric current and drift class 12 physics CBSE

Guru Purnima speech in English in 100 words class 7 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

Copper is not used as potentiometer wire because class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE