# Thread:

1. ## Induction

Is the the correct approach to work through this problem

23n – 1 is divisible by 7 for all n>=1
n(1)
23(1) – 1 = (2*2*2) – 1
= 7

23n – 1 = 7m
n(k+1)
23(k+1) – 1 = 2.23k-1
= 2 * (7m+1) – 1
= 14m + 2 -1
= 14m-1
=7(2m-1)
Where (2m-1) is an int.
So 23n – 1 is divisible by 7.
Therefore, by P.M.I 23n – 1 is divisible by 7 for n>=1.

2. First, this is extremely difficult to read. Please learn some LaTeX.

You are trying to prove $2^{3n} - 1$ is divisible by $7$ for all $n \geq 1$ (I assume $n$ is also an integer)...

Base Step: $n = 1$.

Then $2^{3\cdot 1} - 1 = 8 - 1$

$=7$ which is divisible by $7$.

Inductive Step: Assume this statement is true for $n = k$.

Then $2^{3k} - 1 = 7m$ where $m$ is an integer.

Now let $n = k + 1$ we have

$2^{3(k + 1)} - 1 = 2^{3k + 3} - 1$

$= 2^{3k}\cdot 2^3 - 1$

$= 8\cdot 2^{3k} + 8 - 7$

$= 8(2^{3k} + 1) - 7$

$= 8\cdot 7m - 7$

$= 7(8m - 1)$

which is divisible by $7$.

Therefore $2^{3n}- 1$ is divisible by $7$ for all integer $n \geq 1$.

3. Thanks ProveIt, I completely forgot about Latex for a minute and just pasted in the equation I was working on in word, so sorry!

How does the negative 1 in this line $= 2^{3k}\cdot 2^3 - 1$

change to negative 7?

$=8\cdot 2^{3k} + 8 - 7$

4. Sorry, that should actually read

$8\cdot 2^{3k} - 8 + 7$

$= 8(2^{3k} - 1) + 7$.

The rest follows.

5. ok, thanks
But how/where does the seven come from?

6. $-8 + 7 = -1$.

It just takes a bit of practice, I knew I needed to create a $7$ in order to have this value be divisible by $7$.

7. Please excuse my ignorance, as mathematical Induction and I just aren't getting along.
But did you just place the "7" in the equation because you needed it. It didn't come from any other part of the equation?

8. That's correct. And I also realised I'd need to take out a common factor of $8$.

Like I said, it just comes from experience.
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