Proving $6^n - 1$ is always divisible by 5 by direct proof, not induction

Question

I'm trying to solve the problem:

$5|(6^n-1)$ with $n > 0$

I'm thinking about $6^n$ always ends with 6 as the last digit and subtract $1$ we have $5$ as the last digit of $(6^n - 1)$. A number with $5$ as the last digit is always divisible by $5$.

However, I don't know how to use direct proof to prove that.

Edit: I don't want to use induction proof.

What is "direct proof" for you and in this context? And your idea looks pretty good: show that any power of $;6;$ ends in 6...and continue as you wrote there. This is pretty straightforward....or direct. :) — DonAntonio, Mar 20 '19 at 19:08
This question is attracting so many excellent non-inductive proofs that I recommend it be left open. — Austin Mohr, Mar 20 '19 at 19:29
@DonAntonio With direct proof, I mean write (6^n - 1) equal something as 5k. — John Wang, Mar 20 '19 at 20:53
Every rigorous proof will use induction in one form or another. — Bill Dubuque, Mar 20 '19 at 23:48

score 9 · Answer 1 · answered Mar 20 '19 at 19:10

9

$6^n-1=(6-1)(6^{n-1}+6^{n-2}+\cdots +1)$ and first part is divisible by 5

answered Mar 20 '19 at 19:10

IrbidMath

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Note that this does use induction if you write the proof rigorously. – Bill Dubuque Mar 20 '19 at 23:45

score 3 · Answer 2 · answered Mar 20 '19 at 19:10

3

You have $$6^n-1\equiv 1^n-1=1-1=0\mod 5$$ for every non-negative integer $n$

answered Mar 20 '19 at 19:10

Peter

84,454

score 1 · Answer 3 · answered Mar 20 '19 at 19:10

1

Hint: Expand $6^n=(5+1)^n$ with the binomial theorem.

answered Mar 20 '19 at 19:10

Mike Earnest

75,930

score 0 · Answer 4 · answered Mar 20 '19 at 19:42

If $m$ and $n$ both have $6$ in their unit position, then $mn$ does as well (since $6 \cdot 6 = 36$, leaving a $6$ in the unit position of the product).

Next, wave your hands a bit (or do it formally by induction) to observe that $6 \cdot 6 \cdots 6$ will have a $6$ in its unit position (since you start with $6 \cdot 6 = 36$ and keep multiplying by $6$).

Your argument then follows: Since $6^n$ has a $6$ in its unit position, then $6^n - 1$ has a $5$ in its unit position, and is therefore (by well-known divisibility rules) divisible by $5$.

score 0 · Answer 5 · answered Mar 20 '19 at 19:57

0

For fun:

1)$\sum := 6^{n-1}+6^{n-2}+......6^0$;

2) $6\sum=6^n + 6^{n-1}+.......6^1$;

Subtract: $2)-1)$:

$5\sum= 6^n -6^0=6^n-1$.

Hence?

answered Mar 20 '19 at 19:57

Peter Szilas

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Proving $6^n - 1$ is always divisible by 5 by direct proof, not induction

5 Answers5