TITLE: How many trailing zeroes in $11^{50}-1$?
QUESTION [6 upvotes]: The problem asks us to find how many trailing zeroes there are in decimal representation of $11^{50}-1$. I really don't know how to "attack" this problem. There is a tip that I should try using binomial theorem, but:
$$(1+10)^{50}=\sum_{i=0}^{50}\binom{50}{i}10^i$$
does not look like anything helpful. So how should I use binomial theorem to get something useful?

REPLY [2 votes]: An alternate approach - finding the highest power of $2$ that divides $11^{50}-1$, and then show that the same power of $5$ divides $11^{50}-1$.
Factor $11^{50}-1=(11^{25}-1)(11^{25}+1)$.
$$11^{25}+1\equiv 4\pmod 8\\11^{25}-1\equiv 2\pmod 8$$ so the highest power of $2$ which divides $11^{50}-1$ is $2^3$.
Now you just need to show that $11^{50}-1$ is divisible by $125$. But $11^5\equiv 1\pmod {25}$ and $\frac{11^{50}-1}{11^5-1}=\sum_{k=0}^{9} 11^{5k} \equiv 0\pmod 5$. So $11^{50}-1$ is divisible by $5^3$.
(Why is $11^5-1\equiv 1\pmod {25}$? You can think of it via binomial theorem, $(10+1)^5-1$, or you could think of it as $\mathbb Z_{25}^\times$ being cyclic or order $20$, and $11\equiv 1\pmod 5$ shows that $11$ is a fourth power modulo $25$. Or just work out that $9^4\equiv 11\pmod{25}.$)