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exponents

veronica1999

Member
Jun 4, 2012
63
What is the integer X which satisfies

10^x < 1/2 X 3/4 X 5/6 X 7/8 X .......... X 99/100 < 10^(x+1) ?

Could I get some help?
This contest allows the use of calculators but I still can't think of the right approach.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
What is the integer X which satisfies

10^x < 1/2 X 3/4 X 5/6 X 7/8 X .......... X 99/100 < 10^(x+1) ?

Could I get some help?
This contest allows the use of calculators but I still can't think of the right approach.
Hi Veronica,

This "contest" question is from a previous contest, correct? If it's for an ongoing one we can't help you. Just want to be sure!

Jameson
 

veronica1999

Member
Jun 4, 2012
63
Hi Veronica,

This "contest" question is from a previous contest, correct? If it's for an ongoing one we can't help you. Just want to be sure!

Jameson

I think it's from 10 years ago.:)
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
I am not sure if this will help but what I can think of is to write the middle term as a product.

The top terms are odd so $2n - 1$ and the bottom terms are even so $2n$

Then you have $10^x<\prod\limits_{n=1}^{50}\left(1-\frac{1}{2n}\right)<10^{x+1}$

You can see if that made it any easier.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
What is the integer X which satisfies

10^x < 1/2 X 3/4 X 5/6 X 7/8 X .......... X 99/100 < 10^(x+1) ?

Could I get some help?
This contest allows the use of calculators but I still can't think of the right approach.
Hi veronica1999, :)

\begin{eqnarray}

\frac{1}{2}\times\frac{3}{4}\times\ldots\times \frac{99}{100}&=&\frac{1.2.3.4\cdots99.100}{(2.4.6\cdots.100)^2}\\

&=&\frac{100!}{[2^{50}(1.2.3.\cdots.50)]^2}\\

&=&\frac{100!}{2^{100}(50!)^2}\\

\end{eqnarray}

Assuming that you have a very powerful calculator, :) (The maximum factorial expression that can be computed in my one is about \(70!\))

\[\frac{1}{2}\times\frac{3}{4}\times\ldots\times \frac{99}{100}=\frac{100!}{2^{100}(50!)^2}\approx 0.079589\]

Now you should be able to choose a suitable value for \(x\). :)

Kind Regards,
Sudharaka.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
What is the integer $x$ which satisfies $$10^x < \frac12 \times \frac34 \times \frac56 \times 7frac78 \times \cdots \times \frac{99}{100} < 10^{x+1}\ ?$$
Could I get some help?
This contest allows the use of calculators but I still can't think of the right approach.
You want to show that $$\frac1{100} < \frac12 \times \frac34 \times \frac56 \times \frac78 \times \cdots \times \frac{99}{100} < \frac1{10}.$$

One of those inequalities is quite easy. If you shift the numerator of each fraction one place to the left then you get $$\frac12 \times \frac34 \times \frac56 \times \frac78 \times \cdots \times \frac{99}{100} = \frac1{100}\Bigl(\frac32 \times \frac54 \times \frac76 \times \frac98 \times \cdots \times \frac{99}{98}\Bigr) >\frac1{100}$$ (because each fraction in the bracket is greater than 1).

The other inequality seem much less accessible. You can start by calculating the product of the first few fractions, to get $$\frac12 \times \frac34 \times \frac56 \times \frac78 \times \frac9{10} \approx 0.2460937.$$ But that process soon begins to get tedious. You can shorten it to get something mangeable on even a basic calculator, as follows. Group the next five terms together to form an estimate of their product in this way: $$\frac{11}{12} \times \frac{13}{14} \times \frac{15}{16} \times \frac{17}{18} \times \frac{19}{20} < \Bigl(\frac{19}{20}\Bigr)^5 \approx 0.7737809.$$

Similar calculations for the remaining batches of five fractions (those with denominators 22 to 30, 32 to 40, and so on) show that their products are less than 0.8440802, 0.8810956 and so on – you do the remaining ones. Finally, multiplying all these estimates together, you can conclude that the product of all fifty fractions is less than 0.1.

Messy, but it works. I don't see any simpler way of doing the computation.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,492
One way to prove $\frac{100!}{2^{100}(50!)^2}<\frac{1}{10}$ is using Stirling's approximation, which says $$\sqrt{2\pi}\ n^{n+1/2}e^{-n} \le n! \le e\ n^{n+1/2}e^{-n}.$$ Therefore,

$$
\begin{aligned}
\frac{100!}{2^{100}(50!)^2}\le &
\frac{e\cdot100^{100+1/2}e^{-100}}{2^{100}\left(\sqrt{2\pi}50^{50+1/2}e^{-50}\right)^2} =\\
&\frac{e\cdot100^{100}10e^{-100}}{2^{100}2\pi\cdot50^{100}50e^{-100}}= \\
&\frac{10e}{100\pi}=\frac{e}{\pi}\frac{1}{10}< \frac{1}{10}
\end{aligned}
$$