official quant thread for cat08

[The Last Romeo]

Originally Posted by raghav507

sure buddy,
thoda sa typing main mushkil hai but still yeh lo...

let the roots are p, q, r, s (only for representational purpose)
from the given eqn we get...
pqrs = e/a
pqr+qrs+rsp+spq = -d/a
pq+qr+rs+sp+rp+qs = c/a
p+q+r+s = -b/a

the given expression is
(1-s1)(1-s2)(1-s3)(1-s4)
= 1 - (s1 + s2 + s3 + s4)
+ (s1s2 + s2s3 + s3s4 + s4s1 + s2s4 + s1s3)
- (s1s2s3 s2s3s4 s3s4s1 s1s2s4)
+ (s1s2s3s4)
= 1 - (-b/a) + (c/a) - (-d/a) + (e/a)
= (a+b+c+d+e)/a

hence the answer...
hope it helps...

Definetly it was of great help. I got a bit confused with s1, s2 etc. anywaz thanks a lot

[linksuresh]

New Problem !

f(x) = sqrt[x+2005*sqrt{x+2005*sqrt(x+2005*sqrt...............

where f(x)>0.

Find the remainder when

f(0)f(2006)f(4014)f(6024)f(8036) is divided by 1000.


I neither have the answer nor the solution.

[venkatramaswamy]

Originally Posted by linksuresh

New Problem !

f(x) = sqrt[x+2005*sqrt{x+2005*sqrt(x+2005*sqrt...............

where f(x)>0.

Find the remainder when

f(0)f(2006)f(4014)f(6024)f(8036) is divided by 1000.


I neither have the answer nor the solution.

Are u sure its not 4012 instead of 4014?

[linksuresh]

Originally Posted by venkatramaswamy

Are u sure its not 4012 instead of 4014?

yes... thats what is printed.... if ur unable to proceed treat it as 4012... may be a typo....
need the method more than the answer !

[rahul_kash]

Hey all,

I am not able solve this by Fermat's theorem

Please tell me how to calculate the value of
2^21 mod 23 ,with steps

[prakharc]

Originally Posted by linksuresh

New Problem !

f(x) = sqrt[x+2005*sqrt{x+2005*sqrt(x+2005*sqrt...............

where f(x)>0.

Find the remainder when

f(0)f(2006)f(4014)f(6024)f(8036) is divided by 1000.


I neither have the answer nor the solution.

f(x) = sqrt[x+2005*sqrt{x+2005*sqrt(x+2005*sqrt...............
f(x)= sqrt[x+2005f(x)]
f(x)^2 = x+2005f(x)

Solving the quadratic f(x) =[2005+sqrt{(2005)^2 + 4x}]/2
Now putting value of x=0,2006,4014.....
we get 2005*2006*2007*2008*2009

Divide numerator and denominator by 10 so we get
remainder when 401*1003*2007*2008*2009 is divided by 100
since only unit's digit will affect the result
therefore tens and units digits of the product are
1*3=3;3*7=21;21*8=168;68*9=612 .....so 12

And since we earlier divided by 10 Hence remainder should be 120 Ans

[prince_chittu]

f(x) = sqrt[x+2005*sqrt{x+2005*sqrt(x+2005*sqrt...............

where f(x)>0.

Find the remainder when

f(0)f(2006)f(4014)f(6024)f(8036) is divided by 1000.

f(x)=sqrt[x + 2005*f(x)]
f(x)^2=x+f2005(x)

x=0 : f(x)=2005
f(0)^2 = 2005 f(0)
f(0)=2005 (since f(x)>0)

x=2006: f(x)=2006
f(2006)^2 - 2005f(2006) - 2006=0
[f(2006)-2006][f(2006)+1]=0
Since f(x)>0 , f(2006)= 2006

x=4014 : f(x)=2007
x=6024 : f(x)=2008
x=8036 : f(x)=2009

thus
remainder = (5*6*7*8*9) mod 1000
= 120

[prakharc]

but product of 2005,2006,2007,2008,2009 will end with a 0(due to a 5 and an even number present) which will cancel out a 0 of 1000 so we actually have to get remainder when divided by 100 so just last two digits

[prince_chittu]

@rahul_kash

2^22 mod 23 =1
using inverse method
2^21 mod 23 =12

Sir , to understand the method read Chinese remainder theorem and fermats little theorem from link below

http://www.pagalguy.com/forum/quanti...cat08-514.html (official quant thread for cat08)

[prakharc]

oooops sorry.....remainder will be 120 ...since we divided by 10 earlier :P