# Show that it is divisible by 7 for all positive integers...

• Sep 22nd 2007, 01:39 PM
Bartimaeus
Show that it is divisible by 7 for all positive integers...
Show that $43^n + 83 * 92 (to the power of) (3n-1)$ is divisible by 7 for all positive integers $n$

Thanks for your time guys, really appreciate it and sorry about the dodgy math after the 92, I'm not very good when it comes to this computer stuff.
• Sep 22nd 2007, 01:51 PM
topsquark
Quote:

Originally Posted by Bartimaeus
Show that $43^n + 83 * 92 (to the power of) (3n-1)$ is divisible by 7 for all positive integers $n$

Thanks for your time guys, really appreciate it and sorry about the dodgy math after the 92, I'm not very good when it comes to this computer stuff.

If you know modular arithmetic, then
$43 \equiv 1~\text{(mod 7)}$

$83 \equiv -1~\text{(mod 7)}$

and
$92 \equiv 1~\text{(mod 7)}$

So
$43^n + 83 \cdot 92^{3n-1} \equiv 1^n + (-1) \cdot 1^{3n - 1} ~\text{(mod 7)}$

$\equiv 1 - 1~\text{(mod 7)} ~ \forall ~n$

$\equiv 0 ~\text{(mod 7)}$

So $43^n + 83 \cdot 92^{3n-1}$ is divisible by 7 for all values of n.

-Dan
• Sep 22nd 2007, 02:03 PM
topsquark
Quote:

Originally Posted by Bartimaeus
Show that $43^n + 83 * 92 (to the power of) (3n-1)$ is divisible by 7 for all positive integers $n$

Thanks for your time guys, really appreciate it and sorry about the dodgy math after the 92, I'm not very good when it comes to this computer stuff.

If you don't know modular arithmetic, then consider n = 1:
$43^1 + 83 \cdot 92^{3 \cdot 1 - 1} = 43 + 83 \cdot 92^2 = 702555$
which is divisible by 7.

Now assume the theorem to be true for some n = k.
Assume that
$43^k + 83 \cdot 92^{3k - 1} = 7x$
where x is some integer.

Consider the case for n = k + 1:
$43^{k + 1} + 83 \cdot 92^{3(k + 1) - 1}$

$= 43 \cdot 43^k + 92^3 \cdot 83 \cdot 92^{3k - 1}$
This should be divisible by 7.

Now, consider again:
$43^k + 83 \cdot 92^{3k - 1} = 7x$

Solving this for $83 \cdot 92^{3k - 1}$ gives:
$83 \cdot 92^{3k - 1} = 7x - 43^k$

Inserting this into the k + 1 expression gives:
$43 \cdot 43^k + 92^3 \cdot (7x - 43^k)$

$= 43 \cdot 43^k + 7 \cdot (92^3x) - 92^3 \cdot 43^k$

$= (43 - 92^3) \cdot 43^k + 7(92^3x)$

The last term is manifestly divisible by 7. Let's look at the first term:
$(43 - 92^3) \cdot 43^k$

The coefficient of the $43^k$ is
$43 - 92^3 = -778645 = 7(-111235)$

So
$43^{k + 1} + 83 \cdot 92^{3(k + 1) - 1} = 7(-111235) \cdot 43^k + 7(92^3x)$
where x is an integer. This is obviously divisible by 7.

Thus $43^n + 83 \cdot 92^{3n - 1}$ is divisible by 7 for n = 1, thus for n = 2, thus etc.

-Dan