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help

Name: help 2008-11-17 10:57



There are 3 different integers bigger than 0 :
"a", "b" and "(a+b)/2".

Is it possible the multiplication a*b*(a+b)/2 equal x^2008 where x is integers bigger than 0 too.

Name: Anonymous 2008-11-17 12:44

>>1
Can't be done. You're black.

Name: Anonymous 2008-11-17 17:48

Yeah sure.

Name: Anonymous 2008-11-17 17:53

yes

Name: Anonymous 2008-11-17 19:01

>>1
The more pressing question is, "Who cares?"

Name: 4tran 2008-11-17 22:09

a=b=x=1
Other than that, I don't think there are any solutions.

Name: Anonymous 2008-11-18 10:24

>>6

It says different integers.

Name: Anonymous 2008-11-18 11:52

>>5
Since there's a '2008' in there, it's from some competition.  It's not from the AIME, USAMO, or IMO, and the Putnam hasn't happened yet, so it's either from some small american competition, in which case is should be easy, or a foreign test, which seems more likely since OP seems to have a weak grasp on english.  I can't find it, though.

Name: 4tran 2008-11-19 2:58

>>7
Good eye; I phail.

Name: Anonymous 2008-11-19 14:29

so that a, b, and the average of a and b are both integers and factors of x^2008, huh?

Name: Anonymous 2008-11-19 20:25

>>10
Looks like it, yup

Name: Anonymous 2008-11-22 3:49

Man this question would have been a lot easier last year.

Name: Anonymous 2008-11-22 10:09

Yes. A0 is 1, B0 is 3. That would make (A0+B0)/2 = 2. I'll call the third one C0. The product of A0, B0, and C0 is 6. This is obviously a factor of 6^2008. So, if we multiply each number by 6^(2007/3) or 6^669, we get A = 6^669, C = 2*6^669, and B = 3*6^669, which would give a product of 6^2008.

Name: Anonymous 2008-11-22 11:28

>>13
What about a0 = 4, b0 = 6, (a0+b0)/2 = 5?

Then a0*b0*(a0+b0)/2 = 120, which is obviously a factor of 120^2008, etc...

This seems a litte bit too trivial for my liking, all we have to do is find two integers with integral average.

Perhaps OP is asking, GIVEN an integer x>0, can we find a and b s.t. (a+b)/2 is an integer, and a*b*(a+b)/2 = x^2008 ?

Name: Anonymous 2008-11-22 14:25

>>13
Nicely done.

*tips hat*

Name: Anonymous 2008-11-22 19:34

>>14
OP isn't asking that.

Name: Anonymous 2008-11-23 8:56

>>16

Then it's completely trivial :s

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