7.1 Q-25

Exercise 7.1 — Question 25

Problem

Use the Binomial Theorem to show that $7^n-1$ is divisible by 6 for every positive integer $n$.

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Key Concept

To test divisibility using the Binomial Theorem, rewrite the base so that one part is the divisor (or a multiple of it) and the other part is 1.

$7=6+1$

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Solution

Expand $7^n$ using the Binomial Theorem with $a=6$, $b=1$:

$7^n=(6+1{)}^n={\sum }_{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})6^r\cdot 1^{n-r}$

Writing out the terms:

$7^n=\left(\genfrac{}{}{0ex}{}{n}{0}\right)(6{)}^0+(\genfrac{}{}{0ex}{}{n}{1})(6{)}^1+\left(\genfrac{}{}{0ex}{}{n}{2}\right)(6{)}^2+\cdots +(\genfrac{}{}{0ex}{}{n}{n})(6{)}^n$

$7^n=1+n\cdot 6+\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot 36+\cdots +6^n$

Every term except the first contains a factor of 6. Therefore:

$7^n=1+6\left[n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)\cdot 6+\cdots +6^{n-1}\right]$

$7^n=1+6k\text{where }k\in {\mathbb{Z}}^{+}$

Subtracting 1 from both sides:

$7^n-1=6k$

Since $7^n-1=6k$ for some positive integer $k$, we conclude:

$\boxed{6\mid (7^n-1)\text{ for all }n\in {\mathbb{Z}}^{+}}$

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General Strategy: Divisibility via Binomial Theorem

To show $a^n-1$ is divisible by $(a-1)$:

1. Write $a=(a-1)+1$
2. Expand $(a-1+1{)}^n$ using the Binomial Theorem
3. All terms except $1^n=1$ contain the factor $(a-1)$
4. Therefore $a^n-1$ is divisible by $(a-1)$

Example: $5^n-1$ is divisible by 4, since $5=4+1$, so $5^n=(4+1{)}^n=1+4k$, giving $5^n-1=4k$.

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Remainder Application

From the result above: $7^n=1+6k$

$\text{Remainder when }{7}^n\text{ is divided by }6=1$

This holds for all positive integers $n$.