Question

If we have a number ${{9}^{6}}-11$ , when divided by 8, would leave a remainder of:
(a) $6$
(b) $1$
(c) $2$
(d) $3$

Answer 1

Hint: For solving this question we will write ${{9}^{6}}-11={{\left( 1+8 \right)}^{6}}-11$ and then, expand the term ${{\left( 1+8 \right)}^{6}}$ by the binomial expansion formula for ${{\left( 1+x \right)}^{n}}$ and try to write ${{9}^{6}}-11$ in simpler form so, that we can find the remainder when it is divided by 8 easily.

Complete step-by-step solution -
Given:
We have to find the value of the remainder when we will divide ${{9}^{6}}-11$ by 8.
Now, before we proceed we should know the following formula:

& {{\left( 1+x \right)}^{n}}=1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+{}^{n}{{C}_{3}}{{x}^{3}}+{}^{n}{{C}_{4}}{{x}^{4}}+{}^{n}{{C}_{5}}{{x}^{5}}+.......................+{}^{n}{{C}_{n-1}}{{x}^{n-1}}+{}^{n}{{C}_{n}}{{x}^{n}}................\left( 1 \right) \\\ & \text{Where, }{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} \\\ \end{aligned}$$ Now, we will use the above formula for solving this question. Now, as we can write ${{9}^{6}}-11={{\left( 1+8 \right)}^{6}}-11$ so, using the formula from equation (1) to expand the term ${{\left( 1+8 \right)}^{6}}$ . Then, $\begin{aligned} & {{9}^{6}}-11={{\left( 1+8 \right)}^{6}}-11 \\\ & \Rightarrow {{9}^{6}}-11=1+{}^{6}{{C}_{1}}\times 8+{}^{6}{{C}_{2}}\times {{8}^{2}}+{}^{6}{{C}_{3}}\times {{8}^{3}}+{}^{6}{{C}_{4}}\times {{8}^{4}}+{}^{6}{{C}_{5}}\times {{8}^{5}}+{}^{6}{{C}_{6}}\times {{8}^{6}}-11 \\\ & \Rightarrow {{9}^{6}}-11=1+6\times 8-11+{}^{6}{{C}_{2}}\times {{8}^{2}}+{}^{6}{{C}_{3}}\times {{8}^{3}}+{}^{6}{{C}_{4}}\times {{8}^{4}}+{}^{6}{{C}_{5}}\times {{8}^{5}}+{}^{6}{{C}_{6}}\times {{8}^{6}} \\\ & \Rightarrow {{9}^{6}}-11=1+48-11+{}^{6}{{C}_{2}}\times {{8}^{2}}+{}^{6}{{C}_{3}}\times {{8}^{3}}+{}^{6}{{C}_{4}}\times {{8}^{4}}+{}^{6}{{C}_{5}}\times {{8}^{5}}+{}^{6}{{C}_{6}}\times {{8}^{6}} \\\ & \Rightarrow {{9}^{6}}-11=38+8\times \left( {}^{6}{{C}_{2}}\times {{8}^{1}}+{}^{6}{{C}_{3}}\times {{8}^{2}}+{}^{6}{{C}_{4}}\times {{8}^{3}}+{}^{6}{{C}_{5}}\times {{8}^{4}}+{}^{6}{{C}_{6}}\times {{8}^{5}} \right) \\\ \end{aligned}$ Now, as we know the value of $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$ is always an integer for $n\ge r$ so, the above equation value of ${}^{6}{{C}_{2}}\times {{8}^{1}}+{}^{6}{{C}_{3}}\times {{8}^{2}}+{}^{6}{{C}_{4}}\times {{8}^{3}}+{}^{6}{{C}_{5}}\times {{8}^{4}}+{}^{6}{{C}_{6}}\times {{8}^{5}}$ will also be an integer and let this value be an integer $k$ . Then, $\begin{aligned} & {{9}^{6}}-11=38+8\times \left( {}^{6}{{C}_{2}}\times {{8}^{1}}+{}^{6}{{C}_{3}}\times {{8}^{2}}+{}^{6}{{C}_{4}}\times {{8}^{3}}+{}^{6}{{C}_{5}}\times {{8}^{4}}+{}^{6}{{C}_{6}}\times {{8}^{5}} \right) \\\ & \Rightarrow {{9}^{6}}-11=38+8k \\\ \end{aligned}$ Now, from the above result, we conclude that we can write ${{9}^{6}}-11=38+8k$ , where $k$ is an integer and the value of the remainder when ${{9}^{6}}-11$ is divided by 8 will be equal to the value of the remainder when $38+8k$ is divided by 8. Now, as we have to find the remainder when $38+8k$ is divided by 8. And as $8k$ will be a multiple of 8 so, the value of remainder when $38+8k$ is divided by 8 will be equal to the value of the remainder when $38$ is divided by 8. Now, we will divide $38$ by 8 and find the remainder. Then, $8\overset{4}{\overline{\left){\begin{aligned} & 38 \\\ & \underline{32} \\\ & \underline{06} \\\ \end{aligned}}\right.}}$ Now, from the above result, we conclude that the value of the remainder when we divide $38$ by $8$ will be $6$ . Thus, the value of the remainder when we will divide ${{9}^{6}}-11$ by 8 will be 6. Hence, (a) is the correct option. Note: Here, the student first understands what is asked in the question and then proceeds in the right direction to get the correct answer quickly and avoid calculating the value of ${{9}^{6}}-11$ . After that, we should use the expansion formula for the ${{\left( 1+x \right)}^{n}}$ correctly with suitable values. Moreover, we should be able to analyse the fact that ${}^{6}{{C}_{2}}\times {{8}^{1}}+{}^{6}{{C}_{3}}\times {{8}^{2}}+{}^{6}{{C}_{4}}\times {{8}^{3}}+{}^{6}{{C}_{5}}\times {{8}^{4}}+{}^{6}{{C}_{6}}\times {{8}^{5}}$ will be an integer and write ${{9}^{6}}-11=38+8k$ and find the remainder when we divide 38 by 8 correctly.