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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.

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closed as off-topic by Martin R, Jean-Claude Arbaut, Yanior Weg, José Carlos Santos, user21820 May 21 at 16:34

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    $\begingroup$ 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? $\endgroup$ – Blue May 20 at 9:13
  • $\begingroup$ I don't know what is wrong with this question. Someone has downvoted. $\endgroup$ – Ramesh May 20 at 9:14
  • $\begingroup$ 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. $\endgroup$ – Ramesh May 20 at 9:20
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    $\begingroup$ @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. $\endgroup$ – samerivertwice May 20 at 9:58
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With simple ineqalities we have:

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

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

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

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  • $\begingroup$ 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}.$ $\endgroup$ – DanielWainfleet May 20 at 16:06
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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: $100!\ge (10^{10})^8 \times (10^2)^{10}=(10^{10})^{10}=10^{100}$.

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  • $\begingroup$ I think 100^10 is 10^20 and not 10^100. $\endgroup$ – Ramesh May 20 at 9:25
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Using $$n!>\bigg(\frac{n}{3}\bigg)^{n}, n>8$$

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

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

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