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Share/Save/Bookmark Login/ Register to Bookmark Topic : "INMO 2010" Started by rudra

rudra

#1 Posted 10:40pm 17-01-10  

INMO 2010

[image]39248122.jpg[/image]
   

b555

#2 Posted 10:48pm 17-01-10  

Re: INMO 2010

question 3:

[hide]http://www.mathlinks.ro/viewtopic.php?p=1745211#1745211[/hide]

(solved by prophet sir)
 Edited on 10:50pm 17-01-10    

Soumik

#3 Posted 10:49pm 17-01-10  

Re: INMO 2010

Pls don't provide the links bfore the qsns are solved here.
   

b555

#4 Posted 10:49pm 17-01-10  

Re: INMO 2010

question 6:
[hide]
http://www.mathlinks.ro/viewtopic.php?t=325459[/hide]

(solved by prophet sir)

question 1:
[hide]
http://www.mathlinks.ro/viewtopic.php?t=325446[/hide]
 Edited on 10:50pm 17-01-10    

Soumik

#5 Posted 1:54pm 18-01-10  

Re: INMO 2010

Number 3 is very easy.

Dividng the given equations give x[p]2[/p]y[p]2[/p]z[p]2[/p]=(x[p]2[/p]-xy+y[p]2[/p])(y[p]2[/p]-yz+z[p]2[/p])(z[p]2[/p]-zx+x[p]2[/p]).

Equality is confirmed by A.M-G.M.
   

theprophet

#6 Posted 2:08pm 18-01-10  

Re: INMO 2010

provided you know something about the signs of x,y,z
   

Soumik

#7 Posted 2:12pm 18-01-10  

Re: INMO 2010

Very similar to the inequality u posted here a few days back.
Basically only possiblity is that only 2 reals can be negative.
So best way is again to proceed by modulus.
   

Soumik

#8 Posted 2:40pm 18-01-10  

Re: INMO 2010

Can anyone confirm my ans to the 2nd one?

Primes and nos. of the form 2p, where p is a prime.
 Edited on 2:41pm 18-01-10    

theprophet

#9 Posted 3:19pm 18-01-10  

Re: INMO 2010

add 8,9 to that list. I've posted a solution at mathlinks
   

jishrockz

#10 Posted 6:06pm 18-01-10  

Re: INMO 2010

can anyone tell me the ans of 4? i got 16 tuples.........

n i got the same eq in 3 as soumik replied........
bt i used the identity (a[p]3[/p]+b[p]3[/p]) and (a[p]3[/p]-b[p]3[/p])
jdeyrockz    

b555

#11 Posted 0:27pm 19-01-10  

Re: INMO 2010

http://www.mathlinks.ro/resources.php?c=78&cid=46&year=2010

this is hwere u can get all the questions and the answers..

hit on the question number to get the solution.
   

jishrockz

#12 Posted 4:48pm 19-01-10  

Re: INMO 2010

where the hell?? i can't find the solution of 4........
jdeyrockz    

jishrockz

#13 Posted 4:50pm 19-01-10  

Re: INMO 2010

in 4, i got the condition that a1=a3=a5 and a2=a4=a6........

then tuples were of form p,q,p,q,p,q where p,q are from {1,2,3,4} ........
jdeyrockz    

b555

#14 Posted 9:21pm 19-01-10  

Re: INMO 2010

its been answered a little later:

see this  q4:

http://www.mathlinks.ro/viewtopic.php?p=1746110&sid=0a2e71e9caa941aacecd0ccbae696798#1746110
   

albert

#15 Posted 6:01pm 26-01-10  

Re: INMO 2010

I got the answer of second question as all primes, twice the primes, 8 and 9.
Did someone else get the same?
   
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