comparism

[ahmedulkavi]

There is a question.Which one is greater?211facterial.110^211 or something like that.How can i solve it.
Is there any shortcut. Please explain it.

[rakeen]

hey, I post something about that. Its here. plz search it. But I never got sured that I was right or wrong!

[*Mahi*]

There is a question.Which one is greater?211facterial.110^211 or something like that.How can i solve it.
Is there any shortcut.

Hey , why don't you use LATEX ?It is very much easy to use.
As to the problem:

Try using $AM-GM$ inequality.

[kamrul2010]

I tried like this...

$ n^2 - x^2 < n^2 $
just put $ n=211 , x=1,2,3,4... $ & then multiply all them!
That should do it!

[ahmedulkavi]

Can you explain it broadly,I cannot understand it.

[Shifat]

Yes Here is the wayt o solve it
you see always $(x+a)(x-a)<x^2$ (simple)
factorial 211 can be explained like this.., $1.2.3.4.5..........211$
or $(1.211).(2.210).(3.209)...........(105.107).106$

as mentioned above $(105.107)<106^2$
$(104.108)<106^2$
........................
.........................
$(1.211)<106^2$
so $1.2.3.4.5.................105.107.........210.211< 106^{210}$
both side 106 multiple kore pai $1.2.3.4....................105.106.107......211<106^{211}$

or $211!<106^{211}<110^{211}$
however 106 or above power e eta prove kora easy, but gets technical when it comes less then 106
Hope it helps.

[Shifat]

suppose I am not sure about the solution of this one
which one is greater $100^{300}$ or $300!$ (neurone abaro onuronon problem)
I tried inthe same method but not sure of that one..

[ataher.sams]

It can be done like -----
211!= 211 x 210 x 209 ...........
= 211 x (211-1) x (211-2) ..........
=\[211^{210}\] -(1+2+3+........209+210)

NOW, simply we are understanding that 211! is bigger than \[166^{211}\]....

[sm.joty]

suppose I am not sure about the solution of this one
which one is greater $100^{300}$ or $300!$ (neurone abaro onuronon problem)
I tried inthe same method but not sure of that one..

শিফাত এটা নিয়ে একটা পোস্ট দাও। কারন আমি নিজেও এটা সমাধান করতে পারি নাই।
আর এটা সমাধান করতে গিয়ে অসমতা বিষয়ে নতুন কিছু সমস্যা জন্ম নিয়েছে।
http://www.matholympiad.org.bd/forum/vi ... 7&start=10

[sakibtanvir]

I will give a very short and easy solution with the help of STIRLING'S APPROXIMATION.It says that,\[In(n!)\approx nIn(n)-n \]
Using this We get,\[300!\approx 1411.135\]
&\[In(100^{300})=300In100\approx 1381.551\]
So,$300!>100^{300}$