Question

What will be the remainder when ${{\left( 19 \right)}^{23}}$ is divided by 18?

Answer 1

Hint: Make ${{\left( 19 \right)}^{23}}$into the form ${{\left( a+b \right)}^{n}}$ and then use binomial expansion to expand. After expansion, divide each term of expansion by 18 and find the remainder. Take ‘a’ and ‘b’ such that they are divisible by 18 or one of them is equal to ‘1’ and the other divisible by 18.

Complete step-by-step answer:
Given dividend $={{\left( 19 \right)}^{23}}$ and divisor $=18$

We try to change ${{\left( 19 \right)}^{23}}$ into the form ${{\left( 1+x \right)}^{23}}$ such that ‘x’ is divisible by 18.

So, ${{\left( 19 \right)}^{23}}$ can be written as ${{\left( 1+18 \right)}^{23}}$.

We know binomial expansion of \[{{\left( a+b \right)}^{n}}{{=}^{n}}{{c}_{0}}{{a}^{n}}{{b}^{0}}{{+}^{n}}{{c}_{1}}{{a}^{n-1}}{{b}^{1}}+{{.....}^{n}}{{c}_{n-1}}{{a}^{1}}{{n}^{n-1}}{{+}^{n}}{{c}_{n}}{{a}^{0}}{{b}^{n}}\]

$\Rightarrow {{\left( 1+18 \right)}^{23}}{{=}^{23}}{{c}_{0}}{{\left( 1 \right)}^{23}}{{\left( 18 \right)}^{0}}{{+}^{23}}{{c}_{0}}{{\left( 1 \right)}^{22}}{{\left( 18 \right)}^{1}}{{+}^{23}}{{c}_{22}}{{\left( 1 \right)}^{1}}{{\left( 18 \right)}^{22}}{{+}^{23}}{{c}_{23}}$

In this expansion, all the terms except the first term contain some powers of 18. So, all the terms are divisible by 18. So, the expansion can be written as;

${{\left( 1+18 \right)}^{23}}{{=}^{23}}{{c}_{0}}\left( 1 \right){{\left( 18 \right)}^{0}}+18{{K}_{1}}+........18{{K}_{n-1}}+18{{K}_{n}}$

Property: If ‘a’ is divisible by ‘b’ and ‘c’ is also divisible by ‘b’ then ${{\left( a+c \right)}^{0}}$ will be also divisible by ‘b’.

The expansion can be written;

$\begin{align}

  & 18{{K}_{1}}+18{{K}_{2}}+........18{{K}_{n-1}}+18{{K}_{n}}=18K \\

 & {{\left( 1+18 \right)}^{23}}=1+18K \\

 & {{K}_{1}},{{K}_{2}}.........{{K}_{n-1}},{{K}_{n}}\ \And \ K\text{ are constants}\text{.} \\

\end{align}$

Dividing both sides by 18, we will get;

$\dfrac{{{\left( 1+18 \right)}^{23}}}{18}=\dfrac{18K+1}{18}$

When $\left( 18K+1 \right)$ is divided by 18, we will get ‘1’ as remainder.

Hence, when ${{\left( 19 \right)}^{23}}$will be divided by 18, we will get remainder 1.

Note: Expansion of ${{\left( 19 \right)}^{23}}$into ${{\left( a+b \right)}^{n}}$ is a bit tricky. In this type of question we try to make $a=1$ and choose b as a multiple of divisor so that we can get the remainder easily as other terms will be divisible by the given divisor.