Find the value of n such that 100^n is just greater than 100!

Question

Or consider the general problem- Find the value of n for which x^n is just greater than x!

I dont know even if it is possible to find the solution or not...

Answer 1

Well, if

$$x^n > x!$$

then we can take the logarithm of both sides without affecting the order:

$$\log(x^n) > \log(x!)$$

We get:

$$n \log(x) > \log(1) + \log(2) + \log(3) + \ldots + \log(x)$$

Let's divide by $\log(x)$ on both sides (if $x=1$ then $\log(x) = 0$ and we can't do this, but the problem also has no solution in that case):

$$n > \log_x(1) + \log_x(2) + \ldots + \log_x(x)$$

We can round up to get an integer:

$$n^\star =\left\lceil \frac{\log(1) + \log(2) + \ldots + \log(x)}{\log(x)}\right\rceil$$

---

We can apply this rule to the case where $x=100$. We find that the answer is:

$$n^\star = 79.$$

We can verify that this makes sense because $100!$ is equal to a 158-digit number (93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000)

and $100^{78}$ is a 157 digit number (a one followed by 2*78=156 zeroes), whereas $100^{79}$ is a 159 digit number, hence larger than 100!.

$$\begin{array}{r|l}\text{number} & \text{how many digits}\\\hline 10^{78}& 157\\10!&158\\10^{79}&159\leftarrow\end{array}$$

Answer 2

Use Stirling's approximation and take the log. $x! \approx (\frac xe)^x \sqrt {2 \pi x}$. Take the base $x$ log and round up.

Answer 3

import math 
n=1
while 100**n < math.factorial(100):
    n+=1
print(n)

For completeness, this python code prints the number $79$.