# Is $10^{100}$ (Googol) bigger than $100!$? [closed]

Is $$10^{100}$$ (Googol) bigger than $$100!$$?

If $$10^{100}$$ is called as Googol, does $$100!$$ have any special name to be called, apart from being called as "100 factorial"?

I ask this question because I get to know about the number $$10^{100}$$ on how big it is more often than $$100!$$. If $$100!$$ is bigger than $$10^{100}$$, then why don't we give more focus to $$100!$$ than the other number? Because for me, $$100!$$ looks simple.

## closed as off-topic by Martin R, Jean-Claude Arbaut, Yanior Weg, José Carlos Santos, user21820May 21 at 16:34

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• Have you given this much thought of your own? For instance, have you even tried writing-out what $10^{100}$ and $100!$ equal? ... not the final values, mind you ... just the factors. Each of them has one hundred factors: the first has a lot of $10$s; the second has a few factors less than $10$, one factor equal to $10$, and many factors greater than $10$. What might this suggest to you? – Blue May 20 at 9:13
• I don't know what is wrong with this question. Someone has downvoted. – Ramesh May 20 at 9:14
• I ask this question because I get to know about the number 10^100 on how big it is more often than 100!. If 100! is bigger than 10^100 means why don't we give much focus to 100! than the other number because for me 100! looks simple. – Ramesh May 20 at 9:20
• @Ramesh ignore the downvoters. It's unkind to downvote without a comment in my opinion. What they should have done IMO is leave a comment that you should show some effort on your own part. Anyway, I hope the answers below help you understand how to think about problems like this. – samerivertwice May 20 at 9:58

With simple ineqalities we have:

$$100!\geq 90^{10}\cdot 80^{10}\cdots 20^{10}\cdot 10^{10}$$

$$100!\geq (9\cdot 8 \cdots 2 \cdot 1)^{10}\cdot 10^{90}$$

$$100!\geq (9!)^{10}\cdot 10^{90}>10^{100}$$

• Nice & simple. And shows $100!$ is "a lot" larger, as in the last line, the ratio of $9!^{10}\cdot 10^{90}$ to $10^{100}$ is $(9!/10)^{10}=(3.6288)^{10}\cdot 10^{40}.$ – DanielWainfleet May 20 at 16:06

Before there was an error on the algebra, as pointed out in the comments. I try to fix the error following the same approach:

$$100!=(1\times..\times 10)\times(11\times..\times 20)\times...\times(91\times..\times 100)=A_1...A_{10}$$

so we estimate $$A_i \ge 10^{10}$$ for $$i=2,..9$$.

Instead we write $$A_1A_{10}=(1\times 100)\times(2\times 99)\times(3 \times 97)\times...\times(10 \times 91)\ge (10^2)^{10}$$.

Combining: $$100!\ge (10^{10})^8 \times (10^2)^{10}=(10^{10})^{10}=10^{100}$$.

• I think 100^10 is 10^20 and not 10^100. – Ramesh May 20 at 9:25

Using $$n!>\bigg(\frac{n}{3}\bigg)^{n}, n>8$$

$$100!>\bigg(\frac{100}{3}\bigg)^{100}>10^{100}$$

• You have a typo in the second formula – Thomas May 20 at 12:40
• I think the solution works fine, but the condition $n>8$ is really necessary for the inequality ? – Thomas May 20 at 13:06