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I have the following problem:

Prove that $6^n-1$ is always divisible by 5 using the binomial expansion of $(5+1)^n$.

How can I do this? I don't know how to begin, as I don't see how the binomial expansion relates to the question. Any help would be appreciated.

$$(5+1)^n-1 =underbracenchoose 05^n+ nchoose 15^n-1+...+ nchoose n-15_=5cdot (....)+1-1 = 5k$$

HINT:
$$(5+1)^n=sum_k=0^n binomnk5^k$$
Which terms of this sum are divisible by $5$?

Guide:

• Treat $a=5$ and $b=1$ in $(a+b)^n$, expand it using binomial expansion. Then subtract it by $1$.




• Then try to factor $5$ out from the remaining term.

$$6^n-1 = (5+1)^n-1$$
$$(5+1)^n = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1$$
$$(5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1+n choose n1-1$$
$n choose n1 = 1$ so $1$ and $-1$ cancel out.
$$implies (5+1)^n-1 = n choose 05^n+n choose 15^n-1+n choose 25^n-2+n choose 35^n-3+...n choose n-15^1$$
All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.