Answer
64.8k+ views
Hint: To solve this question, we will use the concept of the binomial theorem for expansion. We will express 7 as the sum or difference of two numbers whose powers are easier to calculate, and then we will use the binomial theorem for that expansion.
Complete step-by-step solution:
Given, ${7^{98}}$
We have to find the remainder when ${7^{98}}$ is divided by 5.
Let us express 7 as the sum or difference of two numbers whose powers are easier to calculate.
So,
${7^{98}}$ can also be represented as,
$
\Rightarrow {7^{98}} = {\left( {{7^2}} \right)^{49}} \\
\Rightarrow {7^{98}} = {\left( {49} \right)^{49}} \\
\Rightarrow {7^{98}} = {\left( {50 - 1} \right)^{49}} \\
$
So, ${7^{98}} = {\left( {50 - 1} \right)^{49}}$.
Now, we will apply the binomial theorem to expand this.
We know that,
$ \Rightarrow {\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ...... + {}^n{C_{n - 1}}a{b^{n - 1}} + {}^n{C_n}{b^n}.$
Therefore,
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = {}^{49}{C_0}{50^{49}} - {}^{49}{C_1}{50^{48}} + {}^{49}{C_2}{50^{47}} + ...... + {}^{49}{C_{48}}50 - {}^{49}{C_{49}}.$
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = {}^{49}{C_0}{50^{49}} - {}^{49}{C_1}{50^{48}} + {}^{49}{C_2}{50^{47}} + ...... + {}^{49}{C_{48}}50 - 1.$
Here, we can see that all the terms except the last term are multiple of 5.
So, we can assume those terms as 5k and write the above statement as:
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = 5k - 1$, where k is any integer value.
Now, adding and subtracting 5, we will get
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = 5k - 1 + 5 - 5$
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = 5\left( {k - 1} \right) + 4$
$ \Rightarrow {7^{98}} = 5\left( {k - 1} \right) + 4$....................………. (i)
Now, as we know that,
$ \Rightarrow $ Dividend = divisor$ \times $quotient + remainder.
So, by comparing this with equation (i), we will get
Remainder = 4.
Hence, we can see that when we divide ${7^{98}}$ by 5, we will get a remainder of 4.
Note: Whenever we ask this type of question, we have to remember some basic points of the binomial expansion. We will express the given value as the sum or difference of two numbers and then we will apply the binomial theorem. After solving, we will get all the terms except the last term, having multiple of 5. Then we will consider all terms as a multiple of 5 and then we will divide that expanded value by 5 and through this, we will get the required value of the remainder.
Complete step-by-step solution:
Given, ${7^{98}}$
We have to find the remainder when ${7^{98}}$ is divided by 5.
Let us express 7 as the sum or difference of two numbers whose powers are easier to calculate.
So,
${7^{98}}$ can also be represented as,
$
\Rightarrow {7^{98}} = {\left( {{7^2}} \right)^{49}} \\
\Rightarrow {7^{98}} = {\left( {49} \right)^{49}} \\
\Rightarrow {7^{98}} = {\left( {50 - 1} \right)^{49}} \\
$
So, ${7^{98}} = {\left( {50 - 1} \right)^{49}}$.
Now, we will apply the binomial theorem to expand this.
We know that,
$ \Rightarrow {\left( {a + b} \right)^n} = {}^n{C_0}{a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ...... + {}^n{C_{n - 1}}a{b^{n - 1}} + {}^n{C_n}{b^n}.$
Therefore,
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = {}^{49}{C_0}{50^{49}} - {}^{49}{C_1}{50^{48}} + {}^{49}{C_2}{50^{47}} + ...... + {}^{49}{C_{48}}50 - {}^{49}{C_{49}}.$
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = {}^{49}{C_0}{50^{49}} - {}^{49}{C_1}{50^{48}} + {}^{49}{C_2}{50^{47}} + ...... + {}^{49}{C_{48}}50 - 1.$
Here, we can see that all the terms except the last term are multiple of 5.
So, we can assume those terms as 5k and write the above statement as:
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = 5k - 1$, where k is any integer value.
Now, adding and subtracting 5, we will get
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = 5k - 1 + 5 - 5$
$ \Rightarrow {\left( {50 - 1} \right)^{49}} = 5\left( {k - 1} \right) + 4$
$ \Rightarrow {7^{98}} = 5\left( {k - 1} \right) + 4$....................………. (i)
Now, as we know that,
$ \Rightarrow $ Dividend = divisor$ \times $quotient + remainder.
So, by comparing this with equation (i), we will get
Remainder = 4.
Hence, we can see that when we divide ${7^{98}}$ by 5, we will get a remainder of 4.
Note: Whenever we ask this type of question, we have to remember some basic points of the binomial expansion. We will express the given value as the sum or difference of two numbers and then we will apply the binomial theorem. After solving, we will get all the terms except the last term, having multiple of 5. Then we will consider all terms as a multiple of 5 and then we will divide that expanded value by 5 and through this, we will get the required value of the remainder.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)