Question

The set of integers n for which ${{2}^{n}}+1$ is divisible by 7
A. $\left\{ 52n|n\in N \right\}$
B. $\left\{ 77n|n\in N \right\}$
C. $\left\{ {{2}^{n}}-1|n\in N \right\}$
D. None of these

Answer 1

Hint: To solve this question, we should write ${{2}^{n}}+1$ as a sum of product of 7 and a remainder. ${{2}^{n}}+1=7k+x$ where k and x are two parameters. To write ${{2}^{n}}+1$ in the desired form, we should note that $8={{2}^{3}}=7+1$. We should take three cases of n where $n=3y,3y+1,3y+2$ and we should write the binomial expansion of ${{2}^{n}}+1$ and see if we get a remainder as zero. For example, if $n=3y$, we get ${{2}^{3y}}+1={{\left( {{2}^{3}} \right)}^{y}}+1={{\left( 7+1 \right)}^{y}}+1$.

Complete step-by-step answer:
We know that ${{\left( a+b \right)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}........{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
\[{{\left( 7+1 \right)}^{y}}+1={{7}^{y}}+{}^{y}{{C}_{1}}{{7}^{y-1}}{{1}^{1}}+{}^{y}{{C}_{2}}{{7}^{y-2}}{{1}^{2}}........{}^{y}{{C}_{y-1}}{{7}^{1}}{{1}^{y-1}}+{}^{y}{{C}_{y}}{{7}^{0}}{{1}^{y}}+1\].
It can be written as \[{{\left( 7+1 \right)}^{y}}+1=7k+1+1=7k+2\]
So, we can conclude that when n = 3y, there is a remainder of 2. So, ${{2}^{n}}+1$ is not divisible by 7 when n = 3y. Similarly, we have to check for other values of n.

We know the formula
${{\left( a+b \right)}^{n}}={{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}........{}^{n}{{C}_{n-1}}{{a}^{1}}{{b}^{n-1}}+{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$
We know that ${{2}^{3}}=8=7+1$.
According to this equation, we divide the problem into 3 cases,
$n=3y,3y+1,3y+2$
Let us consider $n=3y$
${{2}^{3y}}+1={{\left( {{2}^{3}} \right)}^{y}}+1={{\left( 7+1 \right)}^{y}}+1={{\left( 7+1 \right)}^{y}}+1={{7}^{y}}+{}^{y}{{C}_{1}}{{7}^{y-1}}{{1}^{1}}+{}^{y}{{C}_{2}}{{7}^{y-2}}{{1}^{2}}........{}^{y}{{C}_{y-1}}{{7}^{1}}{{1}^{y-1}}+{}^{y}{{C}_{y}}{{7}^{0}}{{1}^{y}}+1$
We can take 7 common in all the terms in the summation except the last two terms.
${{2}^{n}}+1={{2}^{3y}}+1=7k+1+1=7k+2$ where k is an integer.
We can clearly see that we get a remainder of 2 when $n=3y$ and ${{2}^{n}}+1$ is not divisible by 7.
Let us consider case-2 $n=3y+1$,
$\begin{align}
  & {{2}^{3y+1}}+1=2\times {{\left( {{2}^{3}} \right)}^{y}}+1=2\times {{\left( 7+1 \right)}^{y}}+1 \\
 & \Rightarrow 2\times \left( {{7}^{y}}+{}^{y}{{C}_{1}}{{7}^{y-1}}{{1}^{1}}+{}^{y}{{C}_{2}}{{7}^{y-2}}{{1}^{2}}........{}^{y}{{C}_{y-1}}{{7}^{1}}{{1}^{y-1}}+{}^{y}{{C}_{y}}{{7}^{0}}{{1}^{y}} \right)+1 \\
\end{align}$
We can rewrite it as
${{2}^{3y+1}}+1=2\left( 7c+1 \right)+1=7l+2+1=7l+3$ where c, l is an integer.
We can clearly see that we get a remainder of 3 when $n=3y+1$ and ${{2}^{n}}+1$ is not divisible by 7.
Let us consider case-3 $n=3y+2$,
$\begin{align}
  & {{2}^{3y+2}}+1=4\times {{\left( {{2}^{3}} \right)}^{y}}+1=4\times {{\left( 7+1 \right)}^{y}}+1 \\
 & \Rightarrow 4\times \left( {{7}^{y}}+{}^{y}{{C}_{1}}{{7}^{y-1}}{{1}^{1}}+{}^{y}{{C}_{2}}{{7}^{y-2}}{{1}^{2}}........{}^{y}{{C}_{y-1}}{{7}^{1}}{{1}^{y-1}}+{}^{y}{{C}_{y}}{{7}^{0}}{{1}^{y}} \right)+1 \\
\end{align}$
We can rewrite it as
${{2}^{3y+2}}+1=4\left( 7a+1 \right)+1=7b+4+1=7b+5$ where a, b is an integer.
We can clearly see that we get a remainder of 5 when $n=3y+1$ and ${{2}^{n}}+1$ is not divisible by 7.
From the three cases, we can clearly infer that ${{2}^{n}}+1$ is not divisible by 7 for any of the value of n.
$\therefore $${{2}^{n}}+1$ is not divisible by 7 for any value and the answer is option-D which is none of the above.

Note: Some students make a mistake by considering ${{2}^{n}}+1={{\left( {{2}^{3}} \right)}^{\dfrac{n}{3}}}+1$ and consider n as a multiple of three. This leads to a single case in the question and they will miss the other two cases where $n=3y+1,3y+2$. In our question, they will get the answer as none of these by considering only one case. In a different question like to check if ${{2}^{n}}+3$ is divisible by 7 or not, they will get an answer as ${{2}^{n}}+3$ is not divisible by 7 when n = 3y but by taking other cases, we will get a range of n for which ${{2}^{n}}+3$ is divisible by 7. That is why, we should include all cases in verifying.