| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 27 May 2008 21:39:29 IST
|
|
|
Consider a sequence { } such that 
How many distinct pairs chosen from this sequence have g.c.d.= 6 ?
|
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2008 00:57:08 IST
|
|
|
a1 = 2. a2 = 3, a3 = 7, a4 = 43
Now since a(n+1) = an^2 - an + 1 and after a1, the next terms are odd... so terms will be of the type odd^2 - odd + 1 odd^2 is always odd... and odd^2 + 1 will become even. then even - odd = odd always....
so there won't be any distinct pair with GCD = 6... 2 will be the only even number in the sequence.
|
---------------------------------------------------------------
* Gaurav Ragtah ( aka Artemis Fowl )
* Agent 'G' [sniper] - SD-6 (Alliance of Twelve)
* Your friendly neighborhood spideyunlimited |
this reply: 17 points
(with 3 
in 4 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2008 15:17:51 IST
|
|
|
Yes Spidey unlimited is correct All the multiples of an even number are even Therefore all the multiples of 6 are even and according to the problem all are odd numbers except 2 therefore such ordered pairs are 0.
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2008 19:27:57 IST
|
|
|
Ok !
But can u tell me how many pairs of no would have g.c.d.=i ( i is a natural number ) ( as a fn of i)
Actually I am looking for a general result for any g.c.d.
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 May 2008 19:38:13 IST
|
|
|
once explain gcd=i.........
|
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Glitter Graphics</DIV></TD></TR></TABLE>
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 May 2008 15:43:20 IST
|
|
|
that is find no of pairs as a function of i for which their g.c.d=i
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 29 May 2008 16:20:41 IST
|
|
|
Hint: can you prove that every term in the sequence is prime to every other term of the sequence, i.e. all terms are mutually co-prime?
|
Time wounds all heels |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 30 May 2008 19:20:09 IST
|
|
|
Yes , that is the way I was talking about .
See, we have
................................... (1)
= ( repeatedly using the formula (1) )
=
which means that 
which means the g.c.d. of any two numbers 
|
this reply: 9 points
(with 1 
in 3 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 May 2008 09:36:27 IST
|
|
|
Even without algebraic manipulation, you can see that (mod ) when 
|
Time wounds all heels |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 31 May 2008 21:38:18 IST
|
|
|
HOW ??? ( that is to be shown in fact !!!! )
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 1 Jun 2008 19:07:04 IST
|
|
|

Hence )
i.e 
Now, =
and hence )
In this way, we can prove that when k>n
By the same argument , we can prove that when k<n too.
Thus, the result that when 
|
Time wounds all heels |
this reply: 25 points
(with 5 
in 5 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 5 Jun 2008 21:20:52 IST
|
|
|
hsbhatt sir
u said
".By the same argument , we can prove that when k<n too."
could u plz show it ...i do not think it is possible
|
B.Tech CSE, ISMU |
this reply: 2 points
(with 0 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
![[Post New]](/templates/default/images/icon_minipost_new.gif) 6 Jun 2008 09:17:38 IST
|
|
|
yeah actually you are right as it contradicts the result for k>n.
So, the result has to be reworded as \ \text{when} \ i>j)
and thus  = 1, i\ne j)
Thanks for pointing out the error
|
Time wounds all heels |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|