How to prove divisibility of a number using the binomial expansion?

Question

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.

Answer 1

$$(5+1)^n-1 =\underbrace{{n\choose 0}5^n+ {n\choose 1}5^{n-1}+...+ {n\choose n-1}5}_{=5\cdot (....)}+1-1 = 5k$$

Answer 2

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.

Answer 3

HINT: $$(5+1)^n=\sum_{k=0}^n \binom{n}{k}5^k$$ Which terms of this sum are divisible by $5$?

Answer 4

$$6^n-1 = (5+1)^n-1$$ $$(5+1)^n = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1+{n \choose n}1$$ $$(5+1)^n-1 = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1+{n \choose n}1-1$$ ${n \choose n}1 = 1$ so $1$ and $-1$ cancel out. $$\implies (5+1)^n-1 = {n \choose 0}5^n+{n \choose 1}5^{n-1}+{n \choose 2}5^{n-2}+{n \choose 3}5^{n-3}+...{n \choose n-1}5^1$$ All terms have a factor of $5$. Therefore, $6^n-1$ is divisible by $5$.