Solving the "n" Challenge - Find the Smallest Integer

[The legend]

Homework Statement

This is just a problem i came up with while doing in equations, and recently saw in a book too.

Find the smallest integer n, for which

n! > 10ⁿ

The Attempt at a Solution

The trial method (with help from the computer) yields 25 to be the answer.

After a bit more of discussion, it did strike that 'n' lies between 20 and 30. (not very helpful )

But i believe there should be a definite way to solve this...

Any help appreciated.

 

[chiro]

Homework Statement

This is just a problem i came up with while doing in equations, and recently saw in a book too.

Find the smallest integer n, for which

n! > 10ⁿ

The Attempt at a Solution

The trial method (with help from the computer) yields 25 to be the answer.

After a bit more of discussion, it did strike that 'n' lies between 20 and 30. (not very helpful )

But i believe there should be a definite way to solve this...

Any help appreciated.

I guess one way that you can approach this is to find the derivative of the two functions in question. You can use the definition of the gamma function to find the derivative of your factorial, and the other functions derivative can be found by using a (e^(ln 10))^n transformation and then using simple derivative rules for exponential.

Another thing besides the derivative is to find when the two are equal and most likely the n will be some real number and not an integer or a rational number.

By using the derivative and the number in which the two are equal, you should be able to show with these what the rate of change is after the two are equal and from that show that based on the derivative it will always remain an inequality for a specific domain.

 

[The legend]

Thanks for the reply, chiro.

I found the derivative for 10^n, but for the n! i used wolfram alpha and got this.
http://www.wolframalpha.com/input/?i=derivative+n!

There is a digamma function involved, but i don't know much of it, and couldn't carry on with the further steps you've hinted.

I am trying to learn more about that function and some others involved (i'm more of a self learner on the internet)...but if you have a simpler way, please share.

 

[chiro]

Thanks for the reply, chiro.

I found the derivative for 10^n, but for the n! i used wolfram alpha and got this.
http://www.wolframalpha.com/input/?i=derivative+n!

There is a digamma function involved, but i don't know much of it, and couldn't carry on with the further steps you've hinted.

I am trying to learn more about that function and some others involved (i'm more of a self learner on the internet)...but if you have a simpler way, please share.

I guess another simpler way is once you find where the two equations are equal, you can use a simple induction argument to assume true for all k > a (a is the lowest integer that inequality is valid) and then show that it is true for k + 1.

So let's say its true for n = a.

Its easy to see that 10^(n+1) = 10 * 10^n and (n+1) * n! is going to be greater in the (n+1)'th term for the factorial since (n+1) > 10 and since all values after some value are true, then this property will guarantee that the inequality holds.

 

[The legend]

I guess another simpler way is once you find where the two equations are equal, you can use a simple induction argument to assume true for all k > a (a is the lowest integer that inequality is valid) and then show that it is true for k + 1.

So let's say its true for n = a.

Its easy to see that 10^(n+1) = 10 * 10^n and (n+1) * n! is going to be greater in the (n+1)'th term for the factorial since (n+1) > 10 and since all values after some value are true, then this property will guarantee that the inequality holds.

But this proof just guarantees the holding of the inequality after n>a.

What the question asks is to find 'a'.

 

[Dick]

You can get a pretty good approximation to the answer by using Stirling's approximation which tells you ln(n!)~n*ln(n)-n and setting it equal to ln(10^n)=n*ln(10). Beyond that approximation I think you need to use fancy functions. It's probably easier to just use your trial method.

 

[The legend]

You can get a pretty good approximation to the answer by using Stirling's approximation which tells you ln(n!)~n*ln(n)-n and setting it equal to ln(10^n)=n*ln(10). Beyond that approximation I think you need to use fancy functions. It's probably easier to just use your trial method.

okay.

thanks a lot for the reply.