CAT 2008: Quantitative Questions a Day 51-100 -The discussions-Brought to you by SMOT

implex

Quote:

Originally Posted by raghav507

he he...may be ur rite...

i actually started from writing the eqn as 2 factors...and then took the individual factors 1000*2, 500*4, 400*5, 250*8, 200*10, 100*20, 50*40...
on reaching 50*40 i saw that this can be further manipulated in the required way...so the solution...
do you have any other better way to solve this ?
and whats the answer...?

well the official solution is pretty similar to yours

they prove that y^2+1 can be 5,10 or 50
and then go on to further prove that only y^2+1=50 gives a positive integral x, and thus xy^2=112
but the solution i gave is a bit work man like !!

I prove that x is even and y is odd
and there after check for 2-3 values which ends up in giving the value , i then prove that a larger value will give not give positive integers!

My solution looks scary, but i always try to use divisibility to find solutions

mahesh_1979

Answer to QQAD 084 is 3 ie option 2..

raghav507

Quote:

Originally Posted by implex

well the official solution is pretty similar to yours

they prove that y^2+1 can be 5,10 or 50
and then go on to further prove that only y^2+1=50 gives a positive integral x, and thus xy^2=112
----------------------
My solution looks scary, but i always try to use divisibility to find solutions

just now , i found out one mistake in my solution...i didn't see that you have mentioned positive integers...and went on to solve the question directly...with result as +-112...

whereas the valid one will be 112 only...

thebytebites

the eqn given is

3x^2-(y+z)x+2(y-z)^2=0

since x<y<z.. therefore we get that minimum value for y-z is -1
so our eqn becomes 3x^2-(y+z)x+2=0

y+z should be the minimum sum of the number havin product as 6
therefore y+z=5... nd this implies y=2, z=3... therefore option (2) 3

there was this another thing i tried, but the answer comes out to be a lil different... i guess there is a mistake in one of the two methods... so jus correct me...

the above eqn can be written as

2(y^2+z^2-2yz)-yx-zx+3x^2
differentiating the above eqn wrt z(for that matter even x or y)

we'll get 4z-4y-x=0
4(z-y)-x=0
min value for z-y=1
therefore x=4 nd z=6..

therefore option (5) none of these..

i know one of them is wrong, so i wanna know WHY they're wrong!!!???

Aarav · 03:31 PM

Quote:

Originally Posted by implex

226) Positive integers x and y satisfy 3x^2-8y^2+3(xy)^2=2008. Find yx^2

Implex 2008 is a BIG number and hence I might have made calculation error, as the equation fails for positive integers x, and y.

Check for my mistake.

x = (-3y^2 + root(9y^4 + 3.4.(8y^2+2008 ))/6

=> D = (3y^2+16)^2 + 3.32.251 - 32.8 => p^2 - 32.745 = a perfect square = q^2, where p is (3y^2+16).
Numbers (32, and 745) being of the form 3k+1, 3k+2 and they will retain this form with (16, 745.2) ..., we will get q to be div by 3 but subtracting 16 from it = 3y^2 doesn't yield any integer y.

Varun Khullar

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Quantitative Question # 084
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Let x, y, z be three distinct integers such that x < y < z and 3x^2 – (y+z)x + 2(y-z)^2 = 0. The minimum positive value of z will be

(1) 2 (2) 3 (3) 5 (4) 8 (5) none of these

differenciating with respect to z

-x -4(y-z) = 0
x+4y =4z
...but this is min of the expression not of z ...

i am stumped..
x ,x+1 ,x+2 ..
.. still trying..

raghav507

Quote:

Originally Posted by Aarav

Implex 2008 is a BIG number and hence I might have made calculation error, as the equation fails for positive integers x, and y.

Check for my mistake.

x = (-3y^2 + root(9y^4 + 3.4.(8y^2+2008 ))/6

=> D = (3y^2+16)^2 + 3.32.251 - 32.8 => p^2 - 32.745 = a perfect square = q^2, where p is (3y^2+16).
Numbers (32, and 745) being of the form 3k+1, 3k+2 and they will retain this form with (16, 745.2) ..., we will get q to be div by 3 but subtracting 16 from it = 3y^2 doesn't yield any integer y.

hi aarav,
i believe that's not the case...i am getting x=4 and y=7 for the equation in my solution mentioned above...
can you please check it for any errors?

Aarav

OK, found the mistake, what I have solved is for 3x^2 - 8y^2 + 3xy^2 = 2008. This doesn't have any integer solutions.

2008 being so large, I assumed mistake in calculation part when it was actually the wrong copying of the problem.

implex

Quote:

Originally Posted by Aarav

OK, found the mistake, what I have solved is for 3x^2 - 8y^2 + 3xy^2 = 2008. This doesn't have any integer solutions.

2008 being so large, I assumed mistake in calculation part when it was actually the wrong copying of the problem.

yupp! for a moment u scared me, if i solved it all wrong !

chasincat

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Quantitative Question # 084
------------------------------------------------------

Let x, y, z be three distinct integers such that x < y < z and 3x^2 – (y+z)x + 2(y-z)^2 = 0. The minimum positive value of z will be

(1) 2 (2) 3 (3) 5 (4) 8 (5) none of these

The equation is a quadratic in x.To have real solutions to the equation the discriminant Should be=>0.
Doing that we get an equation in y and z.

(y+z)^2=>24(Y-Z)^2

In this we need to minimise (Y-Z) so as to make the value of (Y+Z)^2 greater than 24(y-Z)^2 with the least possible values of Yand Z.
The obvious least value of Y-Z is 1.
Using this we get Y+Z= 5 atleast.

(2,3) is the only possible solution set.
(-2,-3) is also a solution set but it can be ruled out since we are asked the least possible positive integer value of z.

Hence 3 is the least possible value of z.

Ans

ption 2

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