Question

Find the remainder when ${3^{91}}$is divided by 80.
(A) 3 (B) 1 (C) 80 (D) 27

Answer 1

Hint: Convert ${3^{91}}$ in the form of ${3^{4k + d}}$ and then put ${3^4} = 81 = \left( {80 + 1} \right)$ and use binomial expansion.

Complete step by step answer:
According to the question, we have to find the remainder when ${3^{91}}$ is divided by 80.
We can write ${3^{91}}$ as:
$
   \Rightarrow {3^{91}} = {3^{88 + 3}} \\
   \Rightarrow {3^{99}} = {3^{88}} \times {3^3} \\
   \Rightarrow {3^{99}} = 27 \times {3^{4 \times 22}} \\
   \Rightarrow {3^{99}} = 27 \times {\left( {{3^4}} \right)^{22}} \\
   \Rightarrow {3^{99}} = 27 \times {\left( {81} \right)^{22}} \\
   \Rightarrow {3^{99}} = 27 \times {\left( {80 + 1} \right)^{22}} \\
$
Now, for ${\left( {80 + 1} \right)^{22}}$, we can use binomial expansion. By doing this we’ll get:
\[
   \Rightarrow {3^{99}} = 27 \times \left( {^{22}{C_0}{{80}^{22}}{ + ^{22}}{C_1}{{80}^{21}} + .....{ + ^{22}}{C_1}{{80}^1}{ + ^{22}}{C_{22}}{{80}^0}} \right) \\
   \Rightarrow {3^{99}} = 27\left( {^{22}{C_0}{{80}^{22}}{ + ^{22}}{C_1}{{80}^{21}} + .....{ + ^{22}}{C_1}{{80}^1} + 1} \right) \\
   \Rightarrow {3^{99}} = 27\left( {^{22}{C_0}{{80}^{22}}{ + ^{22}}{C_1}{{80}^{21}} + .....{ + ^{22}}{C_1}{{80}^1}} \right) + 27 \\
   \Rightarrow {3^{99}} = 27 \times 80\left( {^{22}{C_0}{{80}^{21}}{ + ^{22}}{C_1}{{80}^{20}} + .....{ + ^{22}}{C_1}{{80}^0}} \right) + 27 \\
\]
Now, \[27 \times 80\left( {^{22}{C_0}{{80}^{21}}{ + ^{22}}{C_1}{{80}^{20}} + .....{ + ^{22}}{C_1}{{80}^0}} \right)\] is a multiple of 80. So it can be written as $80n$. So, we have:
\[ \Rightarrow {3^{99}} = 80n + 27\]

Thus we can say that when ${3^{91}}$is divided by 80, it gives remainder 27. Last option is correct.

Note: If we have to find the remainder when a number (let it be D) is divided by another number (let it be d), we try to convert D as $D = dn + k$ in which case k comes out as our remainder.