Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?

My Attempt:

By factoring out powers of $2$, we can write

$$ \begin{align} K &= 2^{50}\times \left(\frac{1} {1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\ &= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\ &= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99} \end{align} $$

How can I solve for $K$ from here?

  • 3
    $\begingroup$ I think there's a $25!$ missing in the numerator $\endgroup$
    – Fabien
    Jul 4, 2014 at 3:27
  • $\begingroup$ @Fabien I agree with this. $\endgroup$
    – MT_
    Jul 4, 2014 at 3:30
  • $\begingroup$ You might want to add to the question whether you specifically want a proof that the calculated value is correct. This would require, if using floating point calculations or any (other) approximation method, a rigorous error estimate placing the value of $K$ between two specific integers. Currently only one of the answers does that, more or less. $\endgroup$ Jul 4, 2014 at 4:07

5 Answers 5


Note that this answer is not completely rigorous, but it was too fun to pass up.

$$K^2 = 101 \cdot \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots \cdot 100 \cdot 100}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots \cdot 99 \cdot 101}$$

Now, $$\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot \dots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot \dots} = \pi / 2$$

(This is known as the Wallis product)

So $K$ is approximately $\sqrt{101 \pi / 2}$ and $\lfloor K \rfloor = 12$

  • 1
    $\begingroup$ I like this one, especially that particular identity for $\pi$. $\endgroup$
    – Yiyuan Lee
    Jul 4, 2014 at 3:40
  • 2
    $\begingroup$ It's not hard to make this rigorous. If $p(n)$ is the $n^{\rm th}$ partial Wallis product, the integral proof on Wikipedia immediately gives $\pi \frac{n}{2n+1} \leq p(n) \leq \frac{\pi}{2}$. In this case, $K^2=101p(50)$, so $50\pi \leq K^2 \leq \frac{101}{2} \pi$, which is easily good enough for our purposes. $\endgroup$
    – Micah
    Jul 4, 2014 at 4:57

Note that $K = \dfrac{2 \cdot 2 \cdot 4 \cdot 4 \cdots 100 \cdot 100}{1 \cdot 2 \cdot 3 \cdot 4 \cdots 99 \cdot 100} = \dfrac{2^{100}(50!)^2}{100!} = \dfrac{2^{100}}{\dbinom{100}{50}}$

It can be shown that the Central Binomial Coefficent satisfies:

$\left(1-\dfrac{1}{8n}\right)\dfrac{2^{2n}}{\sqrt{\pi n}} \le \dbinom{2n}{n} \le \dfrac{2^{2n}}{\sqrt{\pi n}}\left(1-\dfrac{1}{9n}\right)$

for all $n \ge 1$.

Thus, $12.56105 \approx \dfrac{450}{449}\sqrt{50\pi} \le \dfrac{2^{100}}{\dbinom{100}{50}} \le \dfrac{400}{399}\sqrt{50\pi} \approx 12.56456$

Therefore, $\lfloor K \rfloor = 12$.


I'm sure this answer is not rigorous, but perhaps someone can make it so.

By manipulating factorials we get $$\frac{1}{K}=\frac{1}{2^{100}}\binom{100}{50}=P(X\,{=}\,50)\ ,$$ where $X$ is a binomial random variable with $n=100$ and $p=\frac{1}{2}$. Approximating $X$ by a normal random variable $Y$ in the usual way, we have $Y\sim N(50,5^2)$ and so $$\frac{1}{K}\approx P(49.5\,{<}\,Y\,{<}\,50.5)=P(-0.1\,{<}\,Z\,{<}\,0.1) \approx0.0796\ .$$ Hence $$K\approx12.56$$ and so $\lfloor K\rfloor=12$.


I could not get any ideas of getting the answer analytically. So I just wrote the matlab program as below format rat Initial=1 i=1; count=0; while i<100, Initial = Initial*(i+1)/i; i=i+2; count=count+1; a(count+1)= Initial; end;

Got the answer 12 after flooring the end result.


If we make the problem more general and write $$\displaystyle K_n = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{2n}{2n-1}$$ the numerator is $2^n \Gamma (n+1)$ and the denominator is $\frac{2^n \Gamma \left(n+\frac{1}{2}\right)}{\sqrt{\pi }}$. So, $$K_n=\frac{\sqrt{\pi } \Gamma (n+1)}{\Gamma \left(n+\frac{1}{2}\right)}$$

Considering this expression for large values of $n$, we then have $$K_n=\sqrt{\pi } \sqrt{n}+\frac{1}{8} \sqrt{\pi } \sqrt{\frac{1}{n}}+\frac{1}{128} \sqrt{\pi } \left(\frac{1}{n}\right)^{3/2}-\frac{5 \sqrt{\pi } \left(\frac{1}{n}\right)^{5/2}}{1024}+O\left(\left(\frac{1}{n}\right)^3\right)$$ which implies $$\lfloor K_n \rfloor =\lfloor \sqrt{\pi n} \rfloor$$ which is verified for any value of $n \gt 5$.

  • $\begingroup$ Do you mean to imply that among all good rational approximations for $\pi$, there is none with a square numerator? $\endgroup$ Jul 4, 2014 at 6:49

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