[nutan]

Originally Posted by nutan

he people got stuck here can ne body help me out:sad:
i assumed x=ag y=bg
so substituting this into the question i got

g^2*(1-ab) + g*(b) + a = 0

this is a qurdratic and on solving we get g=....... someting
which is a bit big and i am stuck up here
can ne body guide me like am i into the correct approach or ............:crazyeye:

now taking a=b , a=2b, a=3b, a=4b
and substituting g=1,2,3,4,5
we get tomknow that only a few satisfy the above quadratic equation
which will be 4 in number so the answer is CCCCCCCCCCCCCCCCCCCCCCCCCCC

[nutan]

Originally Posted by vallisri

I dint get the approach. Why did u took y=K*g and y=k2*g
please explain.

THATS THE DEFINITION OF GCD
gcd is the greatest common divisor i.e
for 24=2^3 * 3

34=2*17
the gr8est common divisor or simply GCD is 2
another ex GCD(4, is 4

in that way g is the gr8 common divisor so g divides x and y
so g can b written as x=k1*g and y=k2*g got that

[tatimatla]

Originally Posted by vallisri

I dint get the approach. Why did u took y=K*g and y=k2*g
please explain.

since g is the gcd, g is a multiple of both Y and X.
The purpose is to bring the complete equation in a single variable. Better to start off with Y = K*g than X = K*g since the latter gives u a linear equation in x.

[Jaskiran]

Originally Posted by vallisri

I dint get the approach. Why did u took y=K*g and y=k2*g
please explain.

y=K*g u must hav understood y taken...
actually he tati started wid x=k1*g so ||ly y=k2*g
next coz y is only being used in =qn => for simplicity y=k*g has been taken...
AM I rite Tatimatla...

i got d answer by using forming a quad =n as did by nutan n den lukin 4 valuse 2 satisfy it...simply...

I din quite get d other approach shown by Tatimatla..wud like 2 get my doubts cleared by Tati..pls...

[v-factor]

The equation is x + y^2 + g ^ 3 = x*y*g.
Let x= a*g and y=b*g

So, a*g + (b^2)*(g^2) + g^3 = a*b*(g^3)
(1-ab)*(g^2) + (b^2)*g + a = 0

Solving we get the equation for g as
g = (-(b^2) + ((b^4) – 4*a + 4*(a^2)*b) ^ 0.5) / (2 * (1 – (a*b)))

Also g must be > 0.
So the above equation becomes,
(b^4) < (b^2 + 4*(a^2)*b – 4*a)
Simplifying we get (b^2)*(b+1)*(b-1) < 4*a*(a*b -1)

There are only 4 pairs of (a,b) satisfying the above equation – (1,1) , (2,2) , (3,3) , (3,2)
So ans is c) 4


PS: Hope to post a better method soon.:crazyeye:

[krsh.vik]

soln 30)

if g=1
x = y+1 + 2/(y-1)
(x,y) = (5,2) or (5,3)

for all g >1

x = (y^2+g^3)/(gy-1)
let y = gn

x = (g^2*n^2 - n +g^3+n) / (g^2*n-1)
x= n(g^2*n-1) + (g^3+n)/(g^2*n-1)

since x is integer
n<=(g^3+1)/(g^2-1)

g=2
n<=3 hence n=1,3 (x,y) = (4,2) (4,6)

hw to find out whether there are ani more values r not ?

[anurag.one]

hi all
I am not getting qqad after changing my email id
can anyone help

[synapse_123]

hi all..
why cant x=0,y=0 and hence g=0 can be a solution?

thanks
rahul

[krsh.vik]

Originally Posted by v-factor

There are only 4 pairs of (a,b) satisfying the above equation – (1,1) , (2,2) , (3,3) , PS: Hope to post a better method soon.:crazyeye:

x = y will not satisfy the given eqn

[MBA_007]

Originally Posted by synapse_123

hi all..
why cant x=0,y=0 and hence g=0 can be a solution?

thanks
rahul

bcoz question says a pair (x,y) of natural numbers ..........and all natural numbers start from 1