The result must be a multiple of $25.$  $$a=0(\text{mod}25)$$  According to the divisibility rule of $4$, the remainder of the result divided by $4$ is $$1\times 3\times 1\times 3\times \cdots \times 1\times 3\times 1={{3}^{25}}\times 1=3(\text{mod}4)$$.  When the result is $$75(\text{mod}100)$$, it can be divisible by $25$ and have a remainder of $3$ when divided by $4.$ 
What are the last two digits of the result of $$1\times 3\times 5\times 7\times \cdots \times 101$$? 
Options:
1) $$05$$ 
2) $$25$$ 
3) $$50$$ 
4) $$55$$ 
5) $$75$$ 