BdMO National Higher Secondary 2011/4

[BdMO]

Problem 4:
Which one is larger 2011! or, $(1006)^{2011}$? Justify your answer.

[Mehfuj Zahir]

2011!=1.2.3.4.................1006.107.108.................2011
1006^2011=1006.1006.1006.........................1006
Now,considerA/B=1.2.3.4..........1005.1007.1008.1009...........2011/1006^2010
1005.1007=(1006-1)(1006+1)=1006^2-1>1006^2
1004.1008=(1006-2)(1006+2)=1006^2-2^2>1006^2
now prove that B>A

[FahimFerdous]

I've solved it in another way using the AM-GM inequality. Here's the solution:

We know that, the AM of '1 to n' is (n+1)/2 and the GM of '1 to n' is (n!)^(1/n). Now from the AM-GM inequality we can say that, (n+1)/2>=(n!)^(1/n).
We also know that the equality holds iff all the 'n's are equal. But here that's not the case. So, (n+1)/2>(n!)^(1/n)
Or, {(n+1)/2}^n>n!
Now, if we suppose n=2011 then it's solved. So, 1006^2011>2011!

Is my solution correct? :-s

[abir91]

Yes, good job

By the way, you can also do it by induction (proving the AM-GM implicitly I guess )

[Masum]

Well done.The solution by am-gm is smarter

[TIUrmi]

I have done it as Mehfuj did.

[rakeen]

4.এখানে, 2011! = 2011*2010*2009*2008……*1
1006^2011= 1006*1006*……(2011 times)….1006

লখ্য করি,

2011* 2010*….*1006 > 1006^1006
1005*1004*….*1 < 1006^1005


এখন, a>b and c<d ei duita somporke jodi b,d er difference kom hoy and a,c er difference beshi hoy tahole,

ac<bd

so, 2011! < 1006^2011

[Shifat]

I used a basic (x+p)(x-p)<x^2....
2011! = 1.2.3..........2011
=(1.2.3.4.........1005).1006.(1007.1008.1009.........2011)
2011.1<1006^2
2010.2<1006^2
2009.3<1006^2...............so 1005.1007<1006^2
we see now, 1.2.3.4.............1005.1007..........2009.2010.2011<1006^2010
each side we multiply 1006 then we see
LHS<RHS........