Which of the numbers $300!$ and $100^{300}$ is greater

Determine which of the two numbers $300!$ and $100^{300}$ is greater.

My attempt:Since numbers starting from $100$ to $300$ are all greater than $100$. But am not able to justify for numbers between $1$ to $100$

Solution 1:

Take the logarithm of each side, base $100$. Since the $\log$ function is monotonically increasing, this preserves the ordering. $\log_{100}(100^{300})$ is clearly $300$. Meanwhile,

$$\log_{100}300! = \log_{100}1 + \log_{100}2 + \dotsb + \log_{100}300 = \sum_{n=1}^{300} \log_{100}n .$$

We can try using integrals to approximate this series. Since, again, $\log$ is monotonically increasing, we have:

$$\sum_{n=1}^{300} \log_{100}n \geq \int_{1}^{300} \log_{100}x\, dx = \dfrac{1}{\ln 100} \int_{1}^{300} \ln x\, dx = \dfrac{1}{\ln 100} \left[x \ln x - x \right]^{300}_1 = \dfrac{300\ln 300 - 299}{\ln 100} = 306.64... > 300.$$

This implies that $300! \geq 100^{300}.$

Solution 2:

$\displaystyle e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\frac{x^{n+1}}{(n+1)!}+\cdots $

$$e^x>\frac{x^n}{n!}$$ for $x=n$

$$n!>\bigg(\frac{n}{e}\bigg)^n>\bigg(\frac{n}{3}\bigg)^n$$

for $n=300.$ we have $\color{red}{300!>(100)^{300}}$

Solution 3:

$n! > (n/e)^n$ so $300! >(300/e)^{300} > 100^{300} $ since $e < 3$.

To show $n! > (n/e)^n$, it is true for $n=1, 2$, and $3$.

If it is true for $n$ and false for $n+1$, then $n! > (n/e)^n$ and $(n+1)! \le ((n+1)/e)^{n+1}$ so, dividing, $n+1 < \frac{((n+1)/e)^{n+1}}{(n/e)^n} = \frac{(n+1)^{n+1}}{en^n} $ or $e < (1+1/n)^n $ which is false.

Note that this proof can be easily modified to use $e > (1+1/n)^n $ to show that $n! > (n/e)^n$ implies that $(n+1)! > ((n+1)/e)^{n+1}$.

Solution 4:

The easiest way is to collect values in some optimal way.

$$\frac{300!}{100^{300}}=\prod\limits_{k=1}^{300}\frac{k}{100}$$

Let us reduce this by noticing that we can replace $300\cdot 299 \cdot 298 \cdot ... \cdot 292 \cdot 291$ with $290^{10}$ and similarly in each section of 10 by 10 the same switch getting

$$\frac{300!}{100^{300}}=\prod\limits_{k=1}^{300}\frac{k}{100}>\frac{10!}{10^{10}}\prod\limits_{k=1}^{29}\frac{(10k)^{10}}{100^{10}}$$

We can continue simplification:

$$\frac{300!}{100^{300}}>\frac{10!}{10^{10}}\prod\limits_{k=1}^{29}\frac{10^{10}k^{10}}{10^{20}}=\frac{10!}{10^{10}}\prod\limits_{k=1}^{29}\frac{k^{10}}{10^{10}}=\frac{10!(29!)^{10}}{10^{300}}$$

Since we expect a big difference we will ignore $10!$ and continue

$$\frac{10!(29!)^{10}}{10^{300}}>\frac{(29!)^{10}}{10^{300}}$$

We want to prove that $\frac{(29!)^{10}}{10^{300}}> 1$ which means in essence that we expect $$\frac{29!}{10^{30}}>1$$

We can further continue by replacing $29!$ with $(28!!)^2$ of course knowing $29! > (28!!)^2$

$$\frac{29!}{10^{30}}>\frac{(28!!)^2}{10^{30}}$$

This can be reduced to

$$\frac{28!!}{10^{15}}=\frac{2^{14}14!}{10^{15}}$$

Now we just make a series of substitution. Knowing $2^{10}=1024>1000$ we have

$$\frac{2^{14}14!}{10^{15}}>\frac{2^{4}14!}{10^{12}}$$

Well now we have to resort to prime factorization but this is simple since we have it all

$$\frac{2^{4}\cdot 14\cdot 13 \cdot 12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5^{12}2^{24}}$$

It all comes to

$$\frac{2^3\cdot 3^5\cdot 7^2\cdot 11\cdot 13}{5^{10}}$$

We can keep on replacing carefully $11$ with $10$ and $13$ with $12$

$$\frac{2^3\cdot 3^5\cdot 7^2\cdot 11\cdot 13}{5^{10}}>\frac{2^3\cdot 3^5\cdot 7^2\cdot 10\cdot 12}{5^{10}}=\frac{2^6\cdot 3^6 \cdot 7^2}{5^9}$$

Again carefully we replace $7\cdot 3 > 20$

$$\frac{2^6\cdot 3^6 \cdot 7^2}{5^9}>\frac{2^6\cdot 3^4\cdot 20^2}{5^9}=\frac{2^{10}\cdot 3^4}{5^7}$$

Again we have $2^{10}$

$$\frac{2^{10}\cdot 3^4}{5^7}>\frac{1000\cdot 3^4}{5^7}=\frac{2^3\cdot 3^4}{5^4}$$

I think we can handle this now

$$\frac{2^3\cdot 3^4}{5^4}=\frac{648}{625} > 1$$

And this is what we wanted to prove.

All along we have used the fact that these two values are very much away from each other.