Fun math problem!!!





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Discus: College Confidential Café: 2004 Archive: Fun math problem!!!
By Thermodude (Thermodude) on Wednesday, September 08, 2004 - 10:34 pm: Edit

Here's a math problem for fun...I think this one'll be solved fairly quickly:

Which is larger, (2002!)^2 or 2002^2002...

...needs a proof...

By Olivecushion88 (Olivecushion88) on Wednesday, September 08, 2004 - 10:48 pm: Edit

2002*2001*2000*1999... ...*3*2*1 resulted in an overflow error(it used up too much memory to compute) on my ti89, so it replaced it w/ infinity.

whereas 2002^2 =4008004 which is hardly even close to "infinity"

is this sufficient enough as a proof?

By Candi1657 (Candi1657) on Wednesday, September 08, 2004 - 10:50 pm: Edit

Uh...it's 2002^2002...he wouldn't make it THAT simple...

By Thermodude (Thermodude) on Wednesday, September 08, 2004 - 11:14 pm: Edit

lol...actually, what happened was I mistyped it as 2002^2....so I edited it later...but yeah....it is supposed to be 2002^2002.

By Welshie (Welshie) on Wednesday, September 08, 2004 - 11:19 pm: Edit

I'll say 2002^2002 for no other reason than the fact that x^x functions are the fastest increasing.

By Welshie (Welshie) on Wednesday, September 08, 2004 - 11:22 pm: Edit

Ok, well out of curiosity I was just testing that for smaller x values (3,4) and those were counter examples. I think because of that some could use some mathematical induction to show that (x!)^2 is greater than or equal to x^x.

By Welshie (Welshie) on Wednesday, September 08, 2004 - 11:24 pm: Edit

Yeah, I'm going to say (x!)^2 >/= x^x and in this case, (2002!)^2 > 2002^2002. Someone else can use induction to prove it.

By Thermodude (Thermodude) on Wednesday, September 08, 2004 - 11:27 pm: Edit

I do have a proof...but i'll wait a few days before I post it.

By Thermodude (Thermodude) on Wednesday, September 08, 2004 - 11:38 pm: Edit

keep in mind...for X^X and (X!)^2 one of these expressions is not always greater than the other..it depends on the value of X. However, just a note to C'Cer's....since the original problem posted was (2002!)^2 vs 2002^2002...you only need to find a proof for (2002!)^2 vs 2002^2002, not X^X vs (X!)^2.

By Seleucus26 (Seleucus26) on Thursday, September 09, 2004 - 12:06 am: Edit

u ther thermodude?

first disjointed statments:
1st
2002^2002 - 2002! = the summation of 2002i where i is from 1 to 2001.

2nd
assumption: (doesnt matter if its right, couldve chosen either > or <

2002!^2 > 2002^2002

then
2002! > sqrt(2002^2002)
2002! > 2002^1001

do you agree?

By Optimizerdad (Optimizerdad) on Thursday, September 09, 2004 - 06:05 am: Edit

Seleucus:
Neat idea, using sqrt(). You're right, if you can prove 2002! > 2002^1001, it solves the original problem.

There are 2002 terms on the LHS, 1001 on the right. If we group the ones on the LHS as 1001 pairs

(2002)(1)
* (2001)(2)
* (2000)(3)
...
* (1002)(1001)

and compare them to each of the 1001 terms on the right, we have
(2002)(1) = 2002
(2001)(2) > 2002
(2000)(3) > 2002
...
(1002)(1001) > 2002

so the product of the 1001 pairs on LHS > product of the 1001 numbers on RHS. True?

By Thermodude (Thermodude) on Thursday, September 09, 2004 - 06:18 pm: Edit

correct!! Good job, you guys have found the proof. (The proof I had in mind was similar to Optimizerdad except it would be comparing 2002 terms of 2002 to 2002(1), (2001)(2)....(1)(2002)) :-D


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