comparism

[Phlembac Adib Hasan]

Listen, you must avoid using calculators.Because we can find $300! > 100^{300}$ by large computer.So all kinds of calculations must be avoided.

[nafistiham]

and, I think it shouldn't have been
\[In(n!)\approx nIn(n)-n\]
but,
\[ln(n!)\approx nln(n)-n\]

[*Mahi*]

Listen, you must avoid using calculators.Because we can find $300! > 100^{300}$ by large computer.So all kinds of calculations must be avoided.

You can find $ln 100$ approximately without using calculators

I hope you know $ln 10 \approx 2.3$

This gives $ln (100^{300}) \approx 1380$

[Phlembac Adib Hasan]

Many persons have given proofs,but every proof depends on calculators at some steps.BUT IT IS NOT FAIR.The problem asks for such a proof that uses only logic, no use of calculator.
We have to prove $ 300! > 100^{300} $.We can prove a stronger result :\[n!>\left (\frac {n} {3}\right )^n \]for every positive integer.Induction is the easiest way to prove this.

[Phlembac Adib Hasan]

I hope you know $ln 10 \approx 2.3$

Yes, I know it.But this must be avoided in a pure proof.

[Phlembac Adib Hasan]

I found another proof of this problem using Cauchy.

[Phlembac Adib Hasan]

Go to viewtopic.php?f=27&t=1622.