Find largest 2 digit prime factor of number $\frac{ 200! } { (100!)^2}$ .

The answer is 61.

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As the

denominatoris $(100!)^2$ , it has a factor of every two digit number at least two times.So, the prime $\boldsymbol {\mathbb P}$ to be present at least 3 times in the factor of the

numerator 200!we get, ${\mathbb P\leq \frac{200}{3}}$ ${\Rightarrow \mathbb P\leq66.67}$ And the largest prime number smaller than66.67is61.So, the answer will be, $\color{#69047E} {\boxed {61}}$ .