In the Q2 shouldn’t the equation obtained be (n – 48)(m – 12)= 12*48
Also on Factorization R.H.S will be 2^6*3^2
And since they have asked for odd values of n the only possibility of L.H.S would be when the (n – 48) term is odd.
Hence only factors 1, 3, 9 will be considered making corresponding n values – 49, 51, 57.
I found this approach shorter 🙂
Valid point Prateek 🙂 I should have just said “alternative” and not “more efficient” I guess.
The thought behind posting this approach was, if someone does not already know the factorisation approach, coming up with something like that on the spot would be tricky. Yet the answer could still be got logically without need of any very specialised knowledge 🙂 (The first time I encountered such a question, I solved it by this way and it was a couple of years later that I found out that there was a proper algebraic way as well!)
y n cant be 11?
sir, In q2,if it was asked to solve for negative integers as well,how would we have solved?
(from pt. of view of Prateek’s method)
The note in the previous post (algebra – 5) covers “finding all integer solutions”, Shivam. If you want with the limit on n, that also can be extrapolated – the negative solutions for n and all the positive odd ones less than 60. One would have to list out and eliminate, still not too lengthy…
Sir, In q2,if it was asked for negative solns as well, then what would have been the process (from prateek’s soln. point of view)
Sir, In Q 1, (y+40) can’t take values 0,2,3,4,5,8,10,12,15,16,20 and so on all the factors till 40. So should we rule out those values also ?? The answer should not be 36 . If it is so, kindly tell me my mistake.. 🙂 TIA
Sidharth, you’re right, my bad – I solved for y+40 to be a positive integer. There would be around 19 solutions I guess. I have edited it accordingly… thanks for pointing out!
Sir Kindly explain Question 1 , As why we should opt for y+40 > 0 & not 60 – x > 0
Because when we consider y+40>0, 60-x>0 condition has already been taken care of. So (y+40) = 1,2,3,4,5,6…..30,32,40 are not allowed. And (60-x) = 60, 75, 80,…..2400 are not allowed. But when we eliminated 17 cases for (y+40). The cases for (60-x) were automatically removed. Right J?
Sir, in Qn.1 why are those 17 cases being removed? Kindly clarify.
Because we want positive integer solutions. i.e y should be positive. Those 17 cases have (y+40) positive but y negative.
Sir, could you please help me with this question?
How many integral solutions for (x,y) satisfy the equation 4x^2 – 9y^2 = 2100 ?
Express it as (2x+3y)(2x-3y) = 2100. After that apply your brain and you will find the answer, I will not hint more.
sir, in question 1 how we can have total integral solution as 71 please explain
sir, in question 1 how can we have total integral solution as 71 . please explain
That was explained in the previous post; for every positive pair of factors there would be a negative pair.
but sir here positive integral solution is 19 so its total integral solution must be 38.
No, it is twice the roots of 2400 not twice the positive solutions. Here there are also some cases where one is positive and one negative, thanks to the minus sign in the original equation.
sir ,can u please help me in this 2 question .
1.how to find total negative term of (a-b+c)^2003 ? with detail procedure.
2. find the no. of term (1+X^2+x^4)^40 ? with detail explanation .
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