Question

Prove that the ${{12}^{th}}$ power of any number is of the form $13n$ or $13n+1$ .

Answer 1

Hint: For solving this question we will directly apply the result of Fermat’s theorem and then try to prove the result.

Complete step-by-step answer:

Given:
For any number, we have to prove that the 12^th power of any number is of the form $13n$ or $13n+1$ .
Now, before we proceed we should know the result of Fermat’s theorem.
Fermat’s theorem: If $p$ is any prime number and $x$ is another number such that $x$ and $p$ are coprime to each other then, $\left( {{x}^{p-1}}-1 \right)$ is divisible by $p$ . For example: as $x=98$ and $p=11$ are coprime then, $\left( {{98}^{10}}-1 \right)$ is divisible by $11$ .
Now, we will use the result of Fermat’s theorem to prove the result. Let $N$ is any number then, there are two possibilities that either $N$ will be multiple of 13 or not. For each, we will try to find ${{N}^{12}}$ and then analyse it to prove our result.
First case:
When $N$ is not a multiple of 13. Then, HCF of $N$ and 13 will be 1 as 13 is a prime number. Which means $N$ and 13 are coprime to each other. Now, according to Fermat’s theorem when, $x=N$ and $p=13$ . Then, we can say that, $\left( {{N}^{12}}-1 \right)$ will be divisible by 13. Then,
$\begin{align}
  & {{N}^{12}}-1=13n\text{ , where n}\in \text{I }\text{.} \\
 & \Rightarrow {{N}^{12}}=13n+1 \\
\end{align}$
Second case:
 When $N$ is a multiple of 13. Then, we can write that $N=13k$ where $k$ is any integer. Then, ${{N}^{12}}={{\left( 13k \right)}^{12}}=13\times \left( {{13}^{11}}\times {{k}^{12}} \right)=13n$ , where $n$ is any integer.
Thus, from the above results in each case, we can say that for any number $N$ . ${{N}^{12}}$ is of the form $13n$ or $13n+1$ .
Hence, proved.

Note: Here, the student should first try to understand what is asked in the question and proceed in the right direction. Moreover, the student should apply the result of Fermat’s theorem with the correct values to prove the result.