Aggressively Mutliplying Percentages

We can rewrite this in the non-Floor form as follows

$\dfrac { 100 }{ 100 } \dfrac { 200 }{ 100 } \displaystyle \prod _{ n=1 }^{ 99 }{ \left( \dfrac { 100-n }{ 100 } \right) \left( \dfrac { 100+n }{ 100 } \right) } =2\left( \dfrac { 199 }{ 10000 } \right) \prod _{ n=1 }^{ 98 }{ \left( \dfrac { 10000-{ n }^{ 2 } }{ 10000 } \right) }$

which obviously has a value of less than $1$ . Hence the Floor is $0$ .

By Stirling's approximation, $\frac {200!}{100^{200}}$

$=\frac {\sqrt{2\pi*200}(\frac {200}{e})^{200}}{100^{200}}$ $=20 \sqrt{\pi}(\frac {2}{e})^{200}$

Note that $2<e<3$ so $(\frac {2}{e})^{200}$ will tend to 0.

$20\sqrt{\pi}$ is relatively small and will not affect the value much, hence the floored expression is just $\boxed {0}$ .

$\large = \lfloor 1 \% \times 2 \% \times 3 \% \times \cdots \times 199 \% \times 200 \% \rfloor$ $\large =\left\lfloor \dfrac{1}{100} \times \dfrac{2}{100} \times \dfrac{3}{100} \times \cdots \times \dfrac{199}{100} \times \dfrac{200}{100} \right\rfloor$ $\large = \left\lfloor \dfrac{200!}{10^{400}} \right\rfloor = \left\lfloor \dfrac{7.88.. \times 10^{374}}{10^{400}}\right\rfloor$ Because denominator is greater than numerator.Hence, answer will be zero

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Note : $\lfloor x \rfloor = 0 \ \forall \ x \in [0,1)$

Note that for all 1<m<n, (n-m)(n+m) < n^2, which implies that (n-m)(n+m)/n^2 <1. Letting n = 100, we can pair all terms in the product set by letting m vary from 1 to 99, ( $1\% \times 199\%$ , $2\% \times 198\%$ etc). All product pairs are less than 1, and only two terms are left unaccounted for, 100% and 200%, whose product is 2. Clearly the product of all the other pair must be less than 1 since they are all less than 1 individually. So the only question is if the product of the pairs is 0.5 or greater. If even 1 of the product pairs is less than 0.5, then the product of the pairs must also be less than 0.5, and the pair ( $1\% \times 199\%$ ) easily satisfies this, therefore the product of all terms is less than 1, implying that the floor of the total product is zero.

Using logic, we find that there is a larger number of 'divide by 100's, as there are 200 numbers, than there are multiplying by 100 or more. Therefore the floor of the equation is 0.

The non-floor answer to the problem above is estimated at 7.8865787 x10^-26, which is way less than 1. So floor answer is 0.

There are a lot of great solutions already written, but I'll try to do it using math as simple as possible: $.01*.02*.03...*1.99*2$ =

$(.01*2)*(.02*1.99)*(.03*1.98)...*(1*1.01)$

Which is fairly obviously the product of a handful of numbers slightly bigger than one and a lot of numbers between zero and one: a product which is probably close to zero, but if we want to be even more sure, this product also equals

$(.01*2*1*1.01)*(.02*1.99*.99*1.02)*(.03*1.98*.98*1.03)...*(.49*1.50*.50*1.49)$

Intuition, common mathematical patterns, or just running some numbers in our head lead to the conclusion that .49 * .50 * 1.49 * 1.50 has the greatest magnitude of these combinations;

$.49*.50*1.49*1.50≈(\frac{1}{2})^2(\frac{3}{2})^2=\frac{9}{16}$

In other words, our number is the product of 49 numbers between 0 and $\frac{9}{16}$ - the floor of which will certainly be 0.

200!/100^200 Greatest integer = 0

$\frac{200!}{10^{400}}$ has in the denominator $400$ trailling zeroes, and in the numerator less than $100$ trailling zeroes, so $\frac{200!}{10^{400}}<1\rightarrow$ its floor function equals $0$