Finding the minimum value of n

[Andrax]

Find the minimum value of n that satisfies the following equation E($\frac{10^n}{x}$)=2011 where X and n are natural numbers
So first thought was using derivatives we need to obtain an f(n) I can't seem to advance a lot in this exercise
All I got so far is 0<10^n /x - 2011<1 I did something wrong as well giving the original function (this might be completely wrong) f(x) =E($\frac{10^n}{x}$) x-2011x if we use derivatives we obtain the first format, can i like calculate f(10^n) here and then derivate it...?) I'm really confused..

 

[mfb]

What is E()? Based on your attempt, it looks like the floor function.
Why do you want to derive some function?
What about simple testing?

 

[Andrax]

What is E()? Based on your attempt, it looks like the floor function.
Why do you want to derive some function?
What about simple testing?

1E() is the floor function
2yeah makes sense
3smple testing wouldn't get me th minimum value also the numbers are very big I am thinking of giving 10^n / 2012<x<= 10^n /2012 then trying to make x a natural number

 

[mfb]

The n you are looking for is below 10, and for each n there is an easy way to test if it works. The numbers are all below 10 billions, which should be fine for every calculator.
With some clever estimate, it is possible to directly guess the right n.

 

[SammyS]

[ B]Find the minimum value of n that satisfies the following equation $\ \displaystyle \text{floor}\left(\frac{10^n}{x}\right)=2011\$ where X and n are natural numbers .
So first thought was using derivatives we need to obtain an f(n) I can't seem to advance a lot in this exercise .
All I got so far is 0<10^n /x - 2011<1 I did something wrong as well giving the original function (this might be completely wrong) f(x) =E($\frac{10^n}{x}$) x-2011x if we use derivatives we obtain the first format, can i like calculate f(10^n) here and then derivate it...?) I'm really confused..[ /B]

Taking the derivative of the floor function isn't of any help that I can see. The floor function has many discontinuities, and both x and n are natural numbers.

I think you should work a bit more with the inequality

$\displaystyle 0\le\left(\frac{10^n}{x}\right)-2011<1\,,$

which I assume comes from

$\displaystyle 2011\le\left(\frac{10^n}{x}\right)<2012\ .$

Multiply this by x (which is positive).

 

[Andrax]

Thanks Sammy but I've already work it on that didn't get me anywhere this is what I've gotten so far from the base equation we can notice that x must be 10^n/2012<=x<10^n/2011 x is a natural number now this got me n=7 but with only trying can someone find a working way to obtain n just from the above I obtained
Th