Find the last non zero digit of 28!. 
Find the last non zero digit of 28!.

It is very hard to multiply and find the last nonzero digit. I just wanna know that, is there any easy technique to solve this type of problem?
 A: Here is a technique:


*

*Factor all numbers $1,\dots,28$ as a product of twos, fives and other numbers. (Other numbers meaning numbers not divisable by 2 or 5.)

*Calculate the number of fives and remove them, and also remove an equal number of twos. (Each zero at the end is due to a factor of $10=5\times2$, and there are always more twos than fives.)

*For the remaining factors (including the remaining twos), calculate them modulo 10 (simply remove tens) and multiply them. You can also make any other changes modulo 10, like turning a $9$ into $-1$.

*The last digit of this product will be the last nonzero digit of $28!$


Example in the case of $11!$:


*

*$1=1$, $2=2$, $3=3$, $4=2^2$, $5=5$, $6=2\times3$, $7=7$, $8=2^3$, $9=9$, $10=2\times5$, $11=11$.

*There are two fives.

*The remaining factors are $1$, $2$ (6 times), $3$ (twice), $7$, $9$ and $11$.

*The product is $1\times2^6\times3^2\times(-3)\times(-1)\times1=2^63^3=1728$. The last nonzero digit of $11!$ is thus 8. (Check: $11!=39916800$.)


This technique is not optimal, but it works.
Once you figure out why and how it works, you can probably find a way to make it more efficient if you want to do such calculations again.
Note that in the last step it suffices to calculate the product modulo 10, so you can throw away unnecessary tens and hundreds.
A: Perhaps a small reduction in the calculations in other answers.  First, we use the standard technique for finding the power of a prime which divides a factorial:
$$\Bigl\lfloor\frac{28}{5}\Bigr\rfloor+\Bigl\lfloor\frac{28}{25}\Bigr\rfloor
  =5+1=6$$
so $5^6$ is a factor of $28!$ (and $5^7$ is not), and likewise
$$\Bigl\lfloor\frac{28}{2}\Bigr\rfloor+\Bigl\lfloor\frac{28}{4}\Bigr\rfloor
  +\Bigl\lfloor\frac{28}{8}\Bigr\rfloor
  +\Bigl\lfloor\frac{28}{16}\Bigr\rfloor=14+7+3+1=25\ ,$$
so $2^{25}$ is a fcator of $28!$.  Therefore $28!$ ends with $6$ zeros and we need the last digit of $28!/10^6$.  Simplifying modulo $10$, we have
$$2^\equiv2^5\equiv2^9\equiv2^{13}\equiv2^{17}\ ,\quad 3^4\equiv1\ ,\quad
  7^4\equiv1\ ,\quad 9^2\equiv1\ ;$$
cancelling $2$s and $5$s and reducing the resuts modulo $10$ gives
$$\eqalign{28!/10^6
  &\equiv2^{19}1.1.3.1.1.3.7.1.9.1.1.3.3.7.3.1.7.9.9.1.1.1.3.3.1.3.7.7\cr
  &\equiv2^3.3^8.7^5.9^3\cr
  &\equiv8.1.7.9\cr
  &\equiv4\ .\cr}$$
A: Note that
$$5\times 10\times 15\times 20\times 25=5\times(2\cdot 5)\times (3\cdot 5)\times (2^2\cdot 5)\times 5^2=2^3\cdot 3\cdot 5^6$$
and that 
$$10^6=2^6\cdot 5^6=\frac{8\cdot 5\cdot 10\cdot 15\cdot 20\cdot 25}{3}=8\cdot 5\cdot 10\cdot 5\cdot 20\cdot 25.$$
So, in mod $10$, we have
$$\begin{align}\frac{28!}{10^6}&\equiv 1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 9\cdot 1\cdot 2\cdot 3\cdot 4\cdot 3\cdot 6\cdot 7\cdot 8\cdot 9\cdot 1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7\cdot 8\\&\equiv (1\cdot 2\cdot 3\cdot 4\cdot 6\cdot 7)^3\cdot 3\cdot 8^2\cdot 9^2\\&\equiv \{1\cdot 2\cdot 3\cdot 4\cdot (-4)\cdot (-3)\}^3 \cdot 3\cdot 4\cdot 1\\&\equiv 2\cdot 2\\&\equiv 4.\end{align}$$
