jmc

Note that the product of the first $100$ positive odd integers can be written as $1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots 200}{2\cdot 4\cdots 200}=\frac{200!}{2^{100}\cdot 100!}$ Hence, we seek the number of threes in $200!$ decreased by the number of threes in $100!.$ There are $\lfloor \frac{200}{3}\rfloor +\lfloor \frac{200}{9}\rfloor +\lfloor \frac{200}{27}\rfloor +\lfloor \frac{200}{81}\rfloor =66+22+7+2=97$ threes in $200!$ and $\lfloor \frac{100}{3}\rfloor +\lfloor \frac{100}{9}\rfloor +\lfloor \frac{100}{27}\rfloor +\lfloor \frac{100}{81}\rfloor =33+11+3+1=48$ threes in $100!$ Therefore, we have a total of $97-48=\boxed{49}$ threes.