Is it divisible?

Before the proof by induction, perhaps a simpler proof:

We might observe that powers of $6$ always end in $6$ , so the last digit of $6^n - 1$ will always be $5$ , which satisfies the divisibility rule for $5$ .

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$1.$ Show that the base case $n = 1$ is true.

$P(1) \ : \ 6^1 - 1 = 5 \Longrightarrow 5 \mid 5 \ \checkmark$

$2.$ Assume $5 \mid 6^n - 1 \ \forall \ n \in \mathbb{N}$ .

$\displaystyle \begin{aligned} 6^k - 1 & = 5a & \small \color{#3D99F6} \text{Multiply both sides by } 6 \\ 6 \left(6^k - 1\right) & = 6(5a) \\ {\color{#D61F06}{6^{k + 1} - 6}} & = 5(6a) = {\color{#D61F06}{5b}} & \small \color{#3D99F6} \text{Now prove for } n = k + 1 \\ {\color{#20A900}{6^{k + 1} - 1}} & = {\color{#20A900}{5c}} & \small \color{#3D99F6} \text{If } 5 \mid a \text{ and } 5 \mid b \text{, then } 5 \mid a - b \\ \implies {\color{#D61F06}{\left(6^{k + 1} - 1\right)}} - {\color{#20A900}{\left(6^{k + 1} - 6\right)}} & = 5({\color{#D61F06}{b}} - {\color{#20A900}{c}}) = 5d \\ \implies 5 & = 5d \ \checkmark & \small \color{#3D99F6} \text{Thus, it is true for } n = k + 1 \end{aligned}$

$3.$ Since $P(n)$ is true for $n = 1$ and $n = k + 1$ , it is true for all $n$ : $5 \mid 6^n - 1 \ \forall \ n \in \mathbb{N}$ .

$6^n-1 \equiv (5+1)^n - 1 \equiv 1^n - 1 \equiv 0 \text{ (mod 5)}$ $\implies 6^n-1$ is divisible by 5 for any positive integer $n$ .

We know $6^n = 6 \mod 10$

$6^n -1 = 5 \mod 10$

$6^n-1$ always ends with a $5$ as per our proof. So $6^n-1$ is always divisible by 5.

A simple approach may save your time. Here it goes, $6^{n}$ mod 5 will always return 1 as remainder. For the above given question (1-1) mod 5 =0

A simpler proof.. for any positive integer, the last digit of 6^n is 6(n>0) So (6^n-1) must be dividable by 5..

note that 1 is identical to 1^n. So the question becomes 'is (6^n - 1^n) divisible by 5 for all n. now (6^n - 1^n) = (6 - 1) x [6^(n-1) + 6^(n-2) + ........ + 6 + 1). So it is divisible by (6 - 1) i.e. 5! Regards, David

All powers of 6 ands by 6 so $6^n - 1$ always and by 5, and so it is divisible by 5. $\boxed{true}$

We know the cyclicity of $6$ is $1$ .

So whatever power $6$ is raised to, i.e, $6^n$ , will always end with $6$ for all positive integral values of $n$

So $6$ - $1$ = $5$

Hence $6^n -1$ will always end with $5$ and hence divisible by $5$

$6^{n}-1=(6-1)\times(6^{n-1}+6^{n-2}+6^{n-3}+\cdots+6^{2}+6^{1}+1)$

by a telescoping series expansion. Hence $6^{n}-1$ is divisible by (6-1)=5.

Because 6 to any power has a units digit of six (this due to the fact that $6*6$ has this property and can be confirmed with basic modular arithmetic), we can subtract one from the number (leaving us with a units digit of $6-1=5$ which makes the whole number divisible by 5 (the divisibility rule for 5).