How many trailing zeroes in $11^{50}-1$? 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?
 A: Let $A_n=11^n-1$. Then:
$$ A_{2n} = A_n (A_n+2), $$
$$ A_{5n} = A_n (A_n^4+5A_n^3+10A_n^2+10A_n+5 ),$$
from which, given that $A_1=2\cdot 5$:

$$\forall n\geq 2,\qquad  \nu_2(A_{n}) = 2+\nu_2(n),\qquad \nu_5(A_n)=1+\nu_5(n).$$

The previous identities give:
$$\nu_2(A_{50}) = 2+\nu_2(50) = 3,\qquad \nu_5(A_{50})=1+\nu_5(50) = 3,$$
hence the number of trailing zeroes of $A_{50}$ is $\color{red}{3}$.
Here, as usual, $\nu_p(n)$ stands for $\max\{m\in\mathbb{N}:p^m\mid n\}$.
A: The binomial tip can work.
Take $$11^{50}-1=\sum_{i=\mathbf1}^{50}\binom{50}i10^i$$
Observe that the $i$th term has at least $i$ trailing zeros (could be more if the binomial coefficient has trailing zeros too). This means that the $i$th term will not "fix" any of the $i$ rightmost digits obtained in the first $i-1$ terms.
Your job now is to sum $i$ terms until one of your $i$ rightmost digits is not zero. (Hint: that happens quickly)
A: Here's a tedious solution which only uses the binomial theorem. We want to compute the largest $m$ such that:
$$ 11^{50} - 1 \equiv 0 \pmod{10^m}$$
As you noted:
$$ 11^{50} - 1 = \sum_{i=1}^{50} \binom{50}{i} 10^i $$
In order to find this modulo $10^m$, we can ignore every term after the $m-1$th term. That is: 
$$ 11^{50} - 1 \equiv \sum_{i=1}^{m-1} \binom{50}{i} 10^i \pmod{10^m} $$
Time for some arithmetic:
$ \binom{50}{1} 10^1 = 500 $
$$ 500 \equiv 0 \pmod{100} $$
$ \binom{50}{2} 10^2 = 122500 $
$$ 500 + 122500 \equiv 0 \pmod{1000} $$
$ \binom{50}{3} 10^3 = 19600000 $
$$ 500 + 122500 + 19600000 \equiv 3000 \pmod{10000} $$
hence there are 3 trailing zeroes.
A: If you compute the first three terms of the binomial expansion, you get
$$\begin{align}
(1+10)^{50}&=1+50\cdot10+{50\cdot49\over2}100+{50\cdot49\cdot48\over6}1000+(\text{a multiple of }10000)\\
&=1+500+122500+(\text{a multiple of }10000)\\
&=3001+(\text{a multiple of }10000)
\end{align}$$
A: 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 $9^4\equiv 11\pmod{25}.$)
A: Hint: How many trailing zeros are there in
$$\begin{pmatrix}50\\37\end{pmatrix}10^{37}?$$
It is certainly true that the $10^{37}$ has $37$ trailing zeros.  Now, consider $$\frac{50!}{37!13!}.$$
The numerator of this fraction is divisible by $5$, $10$ times and by $25$ twice.  Therefore, the numerator is divisible by $12$ 5's.  $37!$ is divisible by $5$ $7$ times and divisible by $25$ once.  This means that $37!$ is divisible by $8$ $5$'s.  Finally, $13!$ is divisible by $5$ twice.  Therefore, the numerator has $12$ factors of $5$ and the denominator has $10$ factors of $5$.  Therefore,  $$\frac{50!}{37!13!}$$
has 2 trailing zeros (you can check that there are enough $2$'s to make this work).
Now, generalize.
