comparism
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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.
Is there any shortcut. Please explain it.
Re: comparism
hey, I post something about that. Its here. plz search it. But I never got sured that I was right or wrong!
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Re: comparism
Hey , why don't you use LATEX ?It is very much easy to use.ahmedulkavi wrote:There is a question.Which one is greater?211facterial.110^211 or something like that.How can i solve it.
Is there any shortcut.
As to the problem:
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: comparism
I tried like this...
If computers have no doors or fences, who needs Windows and Gates?
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Re: comparism
Can you explain it broadly,I cannot understand it.
Re: comparism
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.
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.
Re: comparism
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..
which one is greater $100^{300}$ or $300!$ (neurone abaro onuronon problem)
I tried inthe same method but not sure of that one..
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Re: comparism
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 \[166^{211}\]....
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 \[166^{211}\]....
Ataher Sams
Re: comparism
শিফাত এটা নিয়ে একটা পোস্ট দাও। কারন আমি নিজেও এটা সমাধান করতে পারি নাই।Shifat wrote: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..
আর এটা সমাধান করতে গিয়ে অসমতা বিষয়ে নতুন কিছু সমস্যা জন্ম নিয়েছে।
http://www.matholympiad.org.bd/forum/vi ... 7&start=10
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........
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Re: comparism
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,$300!>100^{300}$
Using this We get,\[300!\approx 1411.135\]
&\[In(100^{300})=300In100\approx 1381.551\]
So,$300!>100^{300}$
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