[varun nakra1]

Originally Posted by hanuman

Is the ans 15 (option b)?
I used GM<=AM approach

yes 15...

put v=a+b then
ab=(v^3-126)/3v..maximise this to get ab=15.....

frm this v^3=-63 for df(v)/dv=0..then
ab(max)=(-63)^2/3 so (ab)max^3<63^2
thus ab(max)=15

[varun nakra1]

Originally Posted by chetna

Q-> let
P={2,3,4.......,100}
and
Q={101,102........,200}

how many elements of Q are there such they don't have any element of P as a factor?

Ans-i know the answer will be the number of prime numbers between 100 and 200.......but do u guys start counting that during the exams....I know this is stupid...but is there any method to find the number of primes between any set of numbers ?

yaa...do one thing leave out all the numbers ending with (0,2,4,6,8,5) nd look for only those ones which hav their last digits as (1,3,7,9)..now look for primes amongst them...m sure th set gets reduced by a large amount now
m jus kiddin.....i dunno ne methd besides this....

[varun nakra1]

Originally Posted by hanuman

Is the ans 15 (option b)?
I used GM<=AM approach

hanuman....the reason i didn use this approch was that a nd b are real numbers....not necessarily positive...beware..this can be fatal also.....in that case maximise the func(ab) in terms of (a+b) usin second order derivative.....th answer however is th same

[hanuman]

Originally Posted by varun nakra1

hanuman....the reason i didn use this approch was that a nd b are real numbers....not positive...beware..this can be fatal also.....in that case maximise the func(ab) in terms of (a+b) usin second order derivative.....th answer however is th same

I certainly overlooked what u have said.
Sometimes, it really helps when you don't know much

[chetna]

Originally Posted by hanuman

Is the ans 15 (option b)?
I used GM<=AM approach

jai hanuman

absolutely right....ur approach is also good in this case....the ans given by time is sooooo stupid

[chetna]

Originally Posted by varun nakra1

yaa...do one thing leave out all the numbers ending with (0,2,4,6,8,5) nd look for only those ones which hav their last digits as (1,3,7,9)..now look for primes amongst them...m sure th set gets reduced by a large amount now
m jus kiddin.....i dunno ne methd besides this....

thanks varun for this brilliant method.......kabhi strike hi nahi kiya mujhe

[varun nakra1]

Originally Posted by hanuman

I certainly overlooked what u have said.
Sometimes, it really helps when you don't know much

ignorance is bliss but not always.

[chetna]

Originally Posted by varun nakra1

hanuman....the reason i didn use this approch was that a nd b are real numbers....not necessarily positive...beware..this can be fatal also.....in that case maximise the func(ab) in terms of (a+b) usin second order derivative.....th answer however is th same

i didnt get u...whats the problem with that approach.

i got ab <16
now no matter what values of a and b are....ab max will still be 15....

plz tell me

[vinit456]

Originally Posted by chetna

Q-> let
P={2,3,4.......,100}
and
Q={101,102........,200}

how many elements of Q are there such they don't have any element of P as a factor?

Ans-i know the answer will be the number of prime numbers between 100 and 200.......but do u guys start counting that during the exams....I know this is stupid...but is there any method to find the number of primes between any set of numbers ?

I think u can use the euler no here
as it is nothing but no of coprimes to that no

ex e(100) is 25 which means 25 coprimes to 100

e(200) is 80 so bet 100 and 200 we have 80-25=55 primes

e(10)=4 .. so the coprimes are 1,3,7, 5
as even euler theorem is nothing but the offshoot of fermat(p-1) where no raised to the no. of coprimes of the divisor the remainder being 1

Edit:Though havent actually verified the 55 co primes but I think it is correct

[panky82]

Originally Posted by chetna

try this plz and let me know ur approach

Q-a and b are two real numbers such that a^3+b^3=126
and ab is an integer...whats the maximum possible value of ab ?
a)5 b)15 c)6 d)16

I went by options.
It cant be 16: 2 reasons - solving the eqns with ab=16 gives imaginary roots
also for ab to be max a & b shud b as close as poss. and if a=b=4 then values goes>126 , so ab<16.
Try with ab=15. and subts in other eqn it gives real roots.
so b.