[amar_kashyap]

[quote=siddhesh_j]

Originally Posted by Aarav

Since, the norm is to give a question after solving it, here is one from me.
I solved this, but looking for an alternate approach.

Problem #2
Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

lhs= m/n+n/m+1/m=4

= (root(m/n)-root(n/m))^2+1/m=2

clearly , m, n have to be greater than zero , which limits the set of integers to 0<m,n, for real m,n.

now for integers greater than one 1/m is a fraction.

case 1:m=1 then square term =1 , and no integral n exists to satisfy such an eqn.

case 2:m>1 then either of root(m/n) or root(n/m) will be irrational and square of irrational = irrational. to that add 1/m , answer again irrational , while2 is rational.

so lhs= irrational and rhs = rational , hence the proof.

i realise this isnt a gr8 solution ... so shud I still post a problem or not??

anyways here is problem #3

This one a bit easier than above 2:

At what time are the hands of clock together between 2 and 3?
1. 10 10/11 mins past 2
2. 10 10/11 mins to 3
3. 9 1/11 mins past 2
4. 11 1/11 mins past 2

I cud not understand the part of ur solution which assumes/concludes m, and n to be positive integers. Could u elaborate a little please.... i think we should post the new problem only after the solution to the previous problem has been discussed and clarified.....that wud be better ...wht say..

[vijay317]

Originally Posted by siddhesh_j

problem #3

This one a bit easier than above 2:

At what time are the hands of clock together between 2 and 3?
1. 10 10/11 mins past 2
2. 10 10/11 mins to 3
3. 9 1/11 mins past 2
4. 11 1/11 mins past 2

Hmm an easy one Option 2 is the answer.

Hmm at exactly 2pm or a.m angle between hours nd minutes hand is 60 deg.

So say in P mts they are together..
then we have 60 + 0.5P = 6P

per minute - Hours hand covers 1/2 a degree nd minutes hand covers 6 deg.

So we have P = 120/11 = 10 10/11 mins past 2

Prob 4 : DS qstion basically Hope DS qstion is ok in this thread.

Is a+b < c+d if the L.C.M of a and b is equal to HCF of c and d ?

I . a,b,c,d are multiples of 20
II. a+c < b+d

Am not able to solve this fully..??:??:

[varun nakra1]

Hi i am posting prblm#4
If theres a square of each side 'a'..den if we draw a quadrant of a circle wd one of th vertices as a center and th side as the radius, inside th square,i.e. th arc is drawn with in the square and so if we draw 4 such arcs or quadrants with in the square wd all th 4 vertices as centres...den find th area of th portion common to all th 4 quadrants i.e th area where all of dem meet..the figure will b like two leaves intersecting one another with in the square..i hope i hav elucidated th prblm..

[amar_kashyap]

Varun please edit your post, the 4th problem has already been posted and yet to be solved. Solve the 4th problem and then post the new problem. I am not sure about the 2nd question, AARAV bhai problem 2 ka kuch karo

And yeah, Varun you post a new problem only when you have solved the previous problem...I hope that is clear..It keeps things simple and in order..

[vijay317]

Originally Posted by amar_kashyap

Vijay please edit your post, the 4th problem has already been posted and yet to be solved. Solve the 4th problem and then post the new problem. I am not sure about the 2nd question, AARAV bhai problem 2 ka kuch karo

And yeah, Vijay you post a new problem only when you have solved the previous problem...I hope that is clear..It keeps things simple and in order..

Hey I think I solved the 3rd problem nd then only posted the 4th one..and that too I posted it before the second 4th problem is posted.SO who needs to edit whose posts amar??:grab:

[amar_kashyap]

I am sorry vijay, actually the message was meant for Varun.....I have edited my post

[Aarav]

You know what you did here, created a square root and then had put the restrictions.

It's like saying -3 = x = (root(x))^2 = -3 has no solution as the square is always positive.

Originally Posted by siddhesh_j

Problem #2
Prove that: m/n + (n+1)/m = 4 has no solutions in integers.

lhs= m/n+n/m+1/m=4

= (root(m/n)-root(n/m))^2+1/m=2

clearly , m, n have to be greater than zero , which limits the set of integers to 0<m,n, for real m,n.

now for integers greater than one 1/m is a fraction.

[varun nakra1]

Hi
M soo sorry to post it..
Regards
Varun

Originally Posted by amar_kashyap

Varun please edit your post, the 4th problem has already been posted and yet to be solved. Solve the 4th problem and then post the new problem. I am not sure about the 2nd question, AARAV bhai problem 2 ka kuch karo

And yeah, Varun you post a new problem only when you have solved the previous problem...I hope that is clear..It keeps things simple and in order..

[amar_kashyap]

For problem 2. i give one more try:::

m/n+(n+1)/m=4

m^2-4n(m)+n^2+n=0, for n, m not equal to 0

The above equation is quadratic in 'm'.

Solving for m, we have m= 2n+-sqrt(n.(3n-1)),
Assuming that 'n' is an integer, if i can prove that n(3n-1) is never a square, we arrive at the result that 'm' is not an integer...which i think will complete the proof.

Aarav, need ur help in proving this ...n(3n-1) is never a square for any integer value of 'n'....

[chautauqua]

Originally Posted by amar_kashyap

For problem 2. i give one more try:::

m/n+(n+1)/m=4

m^2-4n(m)+n^2+n=0, for n, m not equal to 0

The above equation is quadratic in 'm'.

Solving for m, we have m= 2n+-sqrt(n.(3n-1)),
Assuming that 'n' is an integer, if i can prove that n(3n-1) is never a square, we arrive at the result that 'm' is not an integer...which i think will complete the proof.

Aarav, need ur help in proving this ...n(3n-1) is never a square for any integer value of 'n'....

problem 2:

as shown above, we have to show that n(3n-1) cannot be a square number for any integer n.

The proof has two steps:
1. GCD(n,3n-1) = 1
2. n and 3n-1 cannot be squares of integers simultaneously

1. GCD(n,3n-1) = GCD(n-1,n) (...using euclid's algo) = 1
2. let n=a^2, 3n-1=b^2; 'a' and 'b' are integers

=> 3*a^2 = b^2 + 1

consider 'b' of the form 3k, 3k+1, and 3k-1... In all of the above cases, b^2 + 1 is not divisible by 3. where as the LHS in the above equation is.
Hence prooved!

I hope the above solution makes sense...