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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.

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.

**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

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