200!/(100!)^2

Number Theory Level 4

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

The answer is 61.

1 solution

Md Omur Faruque
Aug 8, 2015

As the denominator is $(100!)^2$ , 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 than 66.67 is 61 .

So, the answer will be, $\color{#69047E} {\boxed {61}}$ .

There is a small mistake. The denominator is $100!^{2}$ not $100!$ .

Tanveer Alam - 5 years, 9 months ago

Thanks dude. Didn't notice that. I've edited my solution accordingly.

MD Omur Faruque - 5 years, 9 months ago

200!/(100!)^2

The answer is 61.

1 solution

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