Chapter 10.7, Problem 26PPS

To determine

To prove that $7^n+5$ is divisible by 6 for all natural numbers n(or to find an counter example to disprove).

Explanation of Solution

Given:

  $7^n+5$

Concept Used:

An identity can be proved using mathematical induction in two steps:

1. First prove the identity is true for $n=1$ .
2. Prove that if it is true for $n=k$ then it is also true for $n=k+1$ .

Calculation:

In order to prove(or disprove) that $7^n+5$ is divisible by 6 for all natural numbers n, first try to prove it using mathematical induction.

First, let’s check if this is true for $n=1$ .

For $n=1$

  $\begin{array}{c}{7}^n+5=7+5\\ =12\end{array}$

And 12 is divisible by 6. Thus, the given identity is true for $n=1$ .

Now, let it is true for $n=k;\text{ for some }k>1$ .

Thus, $\left(7^k+5\right)$ is divisible by 6, i.e.

  $7^k+5=6m$ ; for some integer m.

Now, let’s show that it is also true for $n=k+1$ , i.e. prove that $\left(7^{k+1}+5\right)$ is also divisible by 17.

Since,

  $\begin{array}{c}{7}^{k+1}+5=7^k\cdot 7+5\\ =\left(6m-5\right)\cdot 7+5\left(\begin{array}{l}\because 7^k+5=6m\\ \Rightarrow 7^k=6m-5\end{array}\right)\\ =6m\cdot 7-35+5\\ =6m\cdot 7-30\\ =6\left(7m-5\right)\end{array}$

Thus, $\left(7^{k+1}+5\right)$ is a multiple of 6, i.e. divisible by 6, i.e. the identity is also true for $n=k+1$ , when it is true for $n=k$ .

Thus, by mathematical induction it is proved that $7^n+5$ is divisible by 6 for all natural numbers n.