A number theory problem by Dev Sharma

The inequality $n!>(n/3)^n$ holds for all positive integers $n$ ; apply this to $n=300$ .

Note: One way to prove the inequality is: $\ln(n!)=\sum_{k=2}^n \ln(k)\geq\int_{1}^{n}\ln(x) dx >n(\ln(\frac{n}{e} ))>n(\ln(\frac{n}{3}))=\ln((\frac{n}{3} )^n)$

Another way to prove the inequality is:
We can give a proof by induction on $n$ , assuming that $n!>\frac{n^n}{3^n}$ . Now $(n+1)!>(n+1)\frac{n^n}{3^n}=(\frac{n}{n+1})^n\frac{(n+1)^{n+1}}{3^n}=\frac{1}{(1+1/n)^n}\frac{(n+1)^{n+1}}{3^n}>\frac{(n+1)^{n+1}}{3^{n+1}}$ since $(1+\frac{1}{n})^n<e<3$ . We have in fact shown that $n!>\frac{n^n}{e^n}$ .

okay so what i pretty much did was just picture the expansion in my mind. 100^300 is 300 100s right? and 300! is every number counting down from 300. so in this theres 200 numbers greater than 100 and only 100 numbers less than 100 . so from logical deduction i assume you can figure out that 300! is larger

We can use Stirling's formula to solve the problem quite nicely. For those of you that aren't familiar with the formula I will state it below: $\lim_{n \rightarrow\infty} n! \rightarrow\sqrt{2 \pi n}. ( \frac{n}{e}\ )^n$ With (geometric) convergence from above (i.e. $\ n! > RHS$ for $\ n > 0$ ) $\Rightarrow\ 300! > \sqrt{600 \pi} . (\frac{300}{e})^{300} > 43 \times\ 100^{300 } > 100^{300}$ $\therefore\ 300! > 100^{300}$

I just used a scientific calculator to know the answer. Here is what I did: let x be the geometric mean of first 300 natural numbers, then we can alternatively write 300! = $x^{300}$ so if x > 100 then 300! > $100^{300}$ and vice versa, so the geometric mean x = $\sqrt[300]{300!}$ = 111.758 (appox) so x > 100 therefore, 300! > $100^{300}$

Simplistically:-

1.) log(100^300)= 300 x log(100)=300 x 2=600

2.) log(300!)=log(300)+log(299)+....+log(100)+....+log(10)+.....+log(1)

Log to the base 10 of 100 is 2, of 10 is 1, and of 1 is zero

now pairing: log(300)+log(1) >2
log(299)+log(2)> 2 ...... log(201)+log(100)>2 ..... log(151)+log(150)>2

Therefore all pairs are >2 Therefore log(300!) > 2 x 300 >600

Therefore 300! > 100^300 q.e.d.

The question is to find out the larger number. Okay, If we plug $?$ as relational operator between $300!$ and $100^{300}$ then the desired form is,

(300!) {\color\red{?}} (100^{300}) \Rightarrow {log_{100} (300!)} {\color\red{?}} {300log_{100} (100)} \Rightarrow{ \sum_{x=300}^1 log_{100} x} {\color\red{?}} {300} \Rightarrow { log_{100} (300) + log_{100} (299) + log_{100} (298) \ldots + log_{100} (1) } {\color\red{?}} (300 ) Now, look carefully that, the number of terms in left side is exact to the number 300 in right side.. So, left side = right side will be possible if and only if every term of $log_{100} x$ is $1$ .. But it is not possible. Because $log_{100} (300)$ , $log_{100} (299)$ ....... upto $log_{100} (>100)$ , every term will be $>1$ .. So, $\sum_{x=300}^1 log_{100} x$ > 300. \Rightarrow \boxed{\color\red{300! > 100^{300}}}

We can prove by induction that $n! > 100^n$ for all integers $n \ge 269$ .

Base step: At $n=269$ , we have that $269! \approx 2.467 \times 10^{538} > 1 \times 10^{538} = 10^{269}$ .

Induction step: At $n=k$ (for some $k > 269$ ), assume that $k! > 100^k$ . Then we have that $(k+1)! = k!(k+1) > 100^k(k+1) > 100^k \cdot 270 > 100^k \cdot 100 = 100^{k+1}.$ Since $(k+1)! > 100^{k+1}$ , this completes the induction proof.

Disclaimer: For the base step, I admittedly experimented with trial-and-error on Wolfram Alpha to find that the smallest number $n$ for which $n! > 100^n$ is true is $269$ .

(N/2)^2 > N! > (N/3)^3

Kamenetsky's formula floor(\(\frac{log(2𝜋x300)}{2}\) + 300(log(300)-log(e))) + 1 gives 615 digits in 300! And 300log(100) + 1 gives 601 digits in 100^300. Note that normally to find the number of digits in an exponent the floor function would be used, but in this case, taking the log of this number gives an integer, so floor function is not necessary. 300! > 100^300

300! = 1(300!) = 100^300(.01^300)300! That's greater than 100^300 if 300! has greater than 600 digits.

Number of digits in X = CEILING[log(X)] CEILING[log(N!)] = CEILING[log(N(N-1)...1)] = CEILING[log(N)+log(N-1)+...+1] Number of digits in 300! = CEILING[log(300) +log(299) ... +log(1)]

log(300) = 2.477; log(100) = 2; log(10) = 1. If log(X) was linear, we'd have approximately 300! with 201(2.23) + 89(1.5) digits, which is 448 + 133 = 581 digits. However, log(X) is NOT linear. log(200) = 2.30; log(49) = 1.69. So we have at least 202(2.3) + 89(1.6) digits = 464 + 142 = 606 digits. That's greater than 600. Therefore, (.01^300)300! > 1 and 300! > 100^300.

As per me, $300!\quad =300\quad \times\quad 299\quad \times\quad ...\quad\times\quad2\quad\times\quad 1$ $=\quad 150\times 150\times ...\quad 300\quad times$

$={ 150 }^{ 300 }$ , obviuosly $>{ 100}^{ 300}$

Simple expansion in mind. Imagine that both are expanded. Consider the factorial term, and in that, consider the numbers which are above 100, divide it in such a way that those numbers doesn't go below 100 and you have at least 200 numbers between 1 and 3. Use those to multiply the numbers which are below 100. Now you have 300 numbers which are distinctively more than 100. That means, it's more than the other term.

I arrived at the solution through logical deduction, Both the solutions can be computed though a string of exactly 300 sequential multiplications. Thus the number with the highest mean factor of multiplication must be the largest.

the mean for 300! =(1+300)/2=150.5

while 100^300=100

300!>100^300