Question

What is the remainder if ${7^{23}}$ is divided by $4$ ?

Answer 1

Hint: In the given question, we are required to find the remainder when ${7^{23}}$ is divided by $4$. So, we will use the concepts of binomial theorem in order to find the expansion of ${7^{23}}$. So, we first convert ${7^{23}}$ into a binomial expression by splitting up the term. Then, we convert the number in terms of the factor of $4$ and the left over remainder.

Complete step by step answer:
So, we have, ${7^{23}}$. We split the number $7$ as $\left( {8 - 1} \right)$ and find the value of expression ${7^{23}}$ using the binomial theorem. So, we get,
${7^{23}} = {\left( {8 - 1} \right)^{23}}$
Now, we use the binomial theorem for expanding ${\left( {8 - 1} \right)^{23}}$.
$ \Rightarrow {7^{23}}{ = ^{23}}{C_0}{\left( 8 \right)^{23}}{\left( { - 1} \right)^0}{ + ^{23}}{C_1}{\left( 8 \right)^{22}}{\left( { - 1} \right)^1}{ + ^{23}}{C_2}{\left( 8 \right)^{21}}{\left( { - 1} \right)^2} + {.....^{23}}{C_{23}}{\left( 8 \right)^0}{\left( { - 1} \right)^{23}}$
Now, we see that each term is a multiple of four except the last term. So, we get,
$ \Rightarrow {7^{23}} = 4k{ + ^{23}}{C_{23}}{\left( 8 \right)^0}{\left( { - 1} \right)^{23}}$

Now, substituting the values of ${8^0}$ as one and ${\left( { - 1} \right)^{23}}$ as $ - 1$, we get,
$ \Rightarrow {7^{23}} = 4k - 1\left( {^{23}{C_{23}}} \right)$
So, we find the value of $\left( {^{23}{C_{23}}} \right)$ using the combinations formula $^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$ as $^{23}{C_{23}} = \dfrac{{23!}}{{\left( {23 - 23} \right)!23!}}$.
We know that the value of $0!$ is one. Hence, we get,
$ \Rightarrow {7^{23}} = 4k - \dfrac{{23!}}{{23!}}$

Now, cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow {7^{23}} = 4k - 1$
Now, we know that the value of the remainder cannot be negative. So, we introduce another constant c having value $c = k + 1$.
$ \Rightarrow {7^{23}} = 4c + 4 - 1$
Adding up the constants, we get,
$ \Rightarrow {7^{23}} = 4c + 3$
So, the number ${7^{23}}$ leaves a remainder of $3$ when divided by $4$.

Note:We must be clear with the concepts of binomial theorem in order to tackle these types of problems. We must have a strong grip over simplification rules to get through the given problem. One must take care of calculations while finding the remainder when divided by four to be sure of the final answer.