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Which is greater among $300 !$ and $\sqrt {300^{300}}$ ?

The answer is $300 !$ (my textbook's answer). I do not know how to solve problems involving such large numbers.

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Simplify the second term a little bit (in other words, write $(300^{300})^{1/2}$ as $300^{150}$. Now, expand the terms. – muffle Dec 12 '13 at 7:07
See…. – lhf Dec 12 '13 at 13:54
up vote 48 down vote accepted

HINT: $(300^{300})^{1/2}=300^{150}$, so you’re comparing

$$\underbrace{300\cdot300\cdot\ldots\cdot300}_{150\text{ factors}}$$


$$300\cdot299\cdot\ldots\cdot1=\underbrace{(300\cdot1)\cdot(299\cdot2)\cdot\ldots\cdot(151\cdot150)}_{150\text{ factors}}\;.\tag{1}$$

Show that each of the parenthesized factors in $(1)$ is at least $300$.

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So simple, indeed. Thanks. – Claude Leibovici Dec 12 '13 at 10:14
That was neat. Thanks... – Apurv Dec 12 '13 at 10:27
@Apurv, Claude: You’re welcome. – Brian M. Scott Dec 12 '13 at 20:07
Nice solution........... – juantheron May 30 '14 at 3:56

When calculations involve large factorials, the use of Stirling approximation for $n!$ is very convenient. In your case, you compare $n!$ to $n^{n/2}$. We can take logarithms of both sides and use the fact that Stirling approximation gives $\log(n!)$ close to $n \log(n) - n$; the logarithm of the second term is simply $\dfrac{n \log(n)} {2}$. Using this approximation, you find that the factorial is the largest as soon as $n > 8$. Without any approximation, the factorial is the largest as soon as $n > 2$ [$n=2$ is the only solution of $n! = n^{n/2}$]

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$300!=(300*1) * (299*2) * ... * (151*150)$, 150 pair products total

$(300^{300})^{1/2}= 300^{150} = 300 * 300 * ... * 300$, 150 counts of 300 total

none of pair product in the first line is smaller than $300$, so $300!$ is larger.

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