Question

Using the Binomial theorem, find the remainder when ${5^{103}}$ is divided by 13.

Answer 1

Hint: Try expressing ${5^{103}}$ as $ = 5 \times {\left( {26 - 1} \right)^{51}}$$ = 5 \times {\left( {2 \times 13 - 1} \right)^{51}}$
Now, use the binomial theorem to expand the expression. While finding the remainder, note that all the terms containing 13 as a factor will be cancelled. So the only term left will be the last i.e. the $52^{nd}$ term of the expansion.

Formula used:
Binomial theorem. ${(x + a)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{a^r}} $

Complete step-by-step answer:
We can start by rewriting ${5^{103}}$ in terms of 13.
${5^{103}}$
$ = 5 \times {5^{102}}$
$ = 5 \times {5^{2 \times 51}}$
$ = 5 \times {25^{51}}$
$ = 5 \times {\left( {26 - 1} \right)^{51}}$
$ = 5 \times {\left( {2 \times 13 - 1} \right)^{51}}$
By binomial theorem, we can expand ${\left( {2 \times 13 - 1} \right)^{51}}$ as $\sum\limits_{r = 0}^{51} {{{\left( { - 1} \right)}^r}{}^{51}{C_r}{{26}^{n - r}}} $
i.e. ${\left( {2 \times 13 - 1} \right)^{51}}$= $\sum\limits_{r = 0}^{51} {{{\left( { - 1} \right)}^r}{}^{51}{C_r}{{26}^{51 - r}}} $
Therefore,
$5 \times {\left( {2 \times 13 - 1} \right)^{51}} \equiv 5 \times {}^{51}{C_{51}}{\left( { - 1} \right)^{51}}$ (mod 13) …….since all other terms contain 13 as a factor.
$ \Rightarrow {\text{ }}5 \times {\left( {2 \times 13 - 1} \right)^{51}} \equiv - 5$ (mod 13)
$ \Rightarrow {\text{ }}5 \times {\left( {2 \times 13 - 1} \right)^{51}} \equiv - 5 + 13 \equiv 8$ (mod 13) ……..since remainder cannot be negative.

Hence, the remainder when ${5^{103}}$ is divided by 13, is 8.

Note:
The Binomial theorem is ${(x + a)^n} = \sum\limits_{r = 0}^n {{}^n{C_r}{x^{n - r}}{a^r}} $
Note that, while finding the remainder when divided by 13, all the terms containing 13 as a factor will be cancelled. So the only term left will be the last i.e. the 52^{nd} term of the expansion.