# Divisibility proof

#### idontknow

How to show that $$\displaystyle \; 11^{n} -6\;$$ is divisible by $$\displaystyle 5$$
Without induction

11 = 1 mod 5

1 person

#### idontknow

Just posted to work without induction

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#### topsquark

Math Team
How to show that $$\displaystyle \; 11^{n} -6\;$$ is divisible by $$\displaystyle 5$$
Without induction
This seems to only work for odd n...

-Dan

#### [email protected]

This seems to only work for odd n...

-Dan
It doesn't. It works for all n. Can you check again and perhaps post why you don't think it works? I'm curious whether I did something wrong.

1 person

#### romsek

Math Team
$11^n - 6 \pmod{5} =$

$(2\cdot 5 + 1)^n - 6 \pmod{5} =$

$\left(\sum \limits_{k=0}^n~\dbinom{n}{k}(2\cdot 5)^k\right) - 6 \pmod{5} =$

$1 - 1 \pmod{5} =$

$0 \pmod{5}$

1 person

#### topsquark

Math Team
Pffl!! :furious:

I was using 11n, not $$\displaystyle 11^n$$. Sorry about that. :cry:

-Dan

1 person

#### skipjack

Forum Staff
It works for all n.
No, it works for every non-negative integer $n$.

For such $n$, x - 1 is a factor of the polynomial x$^n$ - 1.

Hence 11$^n$ - 1 is a multiple of 10, and so 11$^n$ - 6 is divisible by 5.

1 person

#### greg1313

Forum Staff
$$\displaystyle (5+6)^n-6\rightarrow5^n\cdots6^n-6\rightarrow5^n-6(6^{n-1}-1)$$

Six to any positive power is a number with six as the last digit. I will leave it to the reader to finish up.

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1 person

#### greg1313

Forum Staff
Of course, one could apply the same sort of idea to $11^n$ ...