Question

The remainder left out when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by 9 is
$\left( a \right)0$
$\left( b \right)2$
$\left( c \right)7$
$\left( d \right)8$

Answer 1

Hint: In this particular question use the concept of binomial expansion of ${\left( {1 + x} \right)^n}$which is given as ${}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + .........$ and write ${8^2} = 64 = \left( {1 + 63} \right)$ and 62 = (63 – 1) so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given equation
${8^{2n}} - {\left( {62} \right)^{2n + 1}}$
Now the above equation is also written as,
$ \Rightarrow {\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$
$ \Rightarrow {\left( {64} \right)^n} - {\left( {62} \right)^{2n + 1}}$
$ \Rightarrow {\left( {1 + 63} \right)^n} - {\left( {63 - 1} \right)^{2n + 1}}$
$ \Rightarrow {\left( {1 + 63} \right)^n} - {\left( { - 1} \right)^{2n + 1}}{\left( {1 - 63} \right)^{2n + 1}}$.................. (1)
Now as we know that ${\left( { - 1} \right)^2} = 1$
$ \Rightarrow {\left( { - 1} \right)^{2n + 1}} = {\left( { - 1} \right)^{2n}}\left( { - 1} \right) = {\left( {{{\left( { - 1} \right)}^2}} \right)^n}\left( { - 1} \right) = - 1{\left( 1 \right)^n} = - 1$ so use this value in equation (1) we have,
$ \Rightarrow {\left( {1 + 63} \right)^n} - \left( { - 1} \right){\left( {1 - 63} \right)^{2n + 1}}$
$ \Rightarrow {\left( {1 + 63} \right)^n} + {\left( {1 - 63} \right)^{2n + 1}}$
Now according to binomial theorem the expansion of ${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + .........$ os use this property in the above equation we have,
$ \Rightarrow \left[ {{}^n{C_0} + {}^n{C_1}\left( {63} \right) + {}^n{C_2}{{\left( {63} \right)}^2} + .........} \right] + \left[ {{}^{2n + 1}{C_0} + {}^{2n + 1}{C_1}\left( { - 63} \right) + {}^{2n + 1}{C_2}{{\left( { - 63} \right)}^2} + .........} \right]$
Now as we know that that ${}^n{C_0} = {}^{2n + 1}{C_0} = 1$, $\left[ {\because {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right]$ so we have,
$ \Rightarrow \left[ {1 + {}^n{C_1}\left( {63} \right) + {}^n{C_2}{{\left( {63} \right)}^2} + .........} \right] + \left[ {1 - {}^{2n + 1}{C_1}\left( {63} \right) + {}^{2n + 1}{C_2}{{\left( {63} \right)}^2} + .........} \right]$
$ \Rightarrow 2 + 63\left[ {\left( {{}^n{C_1} + {}^n{C_2}\left( {63} \right) + ...........} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}\left( {63} \right) + .........} \right)} \right]$
Now we have to find out the remainder when the above equation is divided by 9.
Now as we know that 63 is divisible by 9 seven (7) times, so $63\left[ {\left( {{}^n{C_1} + {}^n{C_2}\left( {63} \right) + ...........} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}\left( {63} \right) + .........} \right)} \right]$ is divisible by 9.
So in $2 + 63\left[ {\left( {{}^n{C_1} + {}^n{C_2}\left( {63} \right) + ...........} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}\left( {63} \right) + .........} \right)} \right]$only 2 is not divisible by 9.
So the remainder is 2 when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by 9.
So this is the required answer.
Hence option (b) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the binomial expansion of ${\left( {1 + x} \right)^n}$ which is stated above, so apply this formula as above applied and arrange its terms we will get the required answer and the important point is to break the given expression in terms of multiple of 9.