Calculate $1\times 3\times 5\times \cdots \times 2013$ last three digits. Calculate $1\times 3\times 5\times \cdots \times 2013$  last three digits.

If $N=1\times 3\times 5\times \cdots \times 2013$
we need to find $N\pmod{1000}$
Now $1000=8\cdot125$
Now, $1\cdot3\cdot5\cdot7\equiv1\pmod 8$  (Find one generalization here ) 
So, $(1\cdot3\cdot5\cdot7)\cdots (2001\cdot2003\cdot2005\cdot2007)\equiv1\cdot1\cdots\cdot1\cdot1\equiv1\pmod 8 $
So, $N\equiv 1\cdot2009\cdot2011\cdot2013\pmod 8\equiv9\cdot11\cdot13\equiv1\cdot3\cdot5\equiv 7\pmod 8$
Clearly, $125|N\implies N\equiv0\pmod{125}$ as $125$ is a factor of $N$
Now applying the Chinese Remainder Theorem,
$$N\equiv 0\cdot b_1\cdot\frac{1000}{125}+7\cdot b_2\cdot\frac{1000}8\pmod{1000}$$
 where  $b_1\cdot\frac{1000}{125}\equiv1\pmod{125}\text{ and } b_2\cdot\frac{1000}8\equiv1\pmod 8$
We don't need to calculate $b_1$ as it's multiplier is $0$
$125b_2\equiv1\pmod8\iff 5b_2\equiv1\pmod8$
Trying multiplication of $5$ with numbers coprime to and $<8,$ we get $b_2\equiv5\pmod 8$
So, $$N\equiv7\cdot5\cdot125\pmod{1000}\equiv(8\cdot4+3)125\equiv3\cdot125\equiv375\pmod{1000}$$
