Fourth Problem for Floors

Algebra Level 4

Find the closed form of the product:

$\left \lfloor \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dots \dfrac{7000}{6999} \right \rfloor$

For more problems like this, try answering this set .

The answer is 104.

1 solution

Christian Daang
Jun 30, 2017

By using the Sterling's Formula, we have the closed product of the product as follows:

$= \left\lfloor \dfrac{2^{3500} \cdot 3500!}{6999!!} \right\rfloor \\ = \left\lfloor \dfrac{2^{7000} \cdot (3500!)^2}{7000!} \right\rfloor \\ = \left\lfloor \dfrac{2^{7000} \cdot \left( \dfrac{3500}{e}\right)^{7000} \cdot 7000\pi }{\left( \dfrac{7000}{e}\right)^{7000} \cdot \sqrt{14000\pi} } \right\rfloor \\ = \boxed{104}$

Fourth Problem for Floors

The answer is 104.

1 solution

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