@chillfactor said:Find all positive integers n for which n ˛ + 89n + 2010 is a perfect square.
sirji fact nhn ho pa rha hai 
sirji fact nhn ho pa rha hai @nick_baba said:sir ye factorize hua hai....(n+44)(n+45)+1iske age kuch clear nai keh sakte...to is it 0??
^^ fact is wrng so avoid it.
(n+44)(n+45)+1 = p^2
(n+44)(n+45) = (p-1)(p+1)
either
p+1 = (n+44)(n+45)
p-1 = 1, p=2
so,3 = (n+44)(n+45) but this is nt possible for +tive in N
p+1 = n+45
p-1 = n+44
2p = 2n + 89
2(p-n) = 89
which is not possible.... 0 soln
@nick_baba said:sir ye factorize hua hai....(n+44)(n+45)+1iske age kuch clear nai keh sakte...to is it 0??but haan -44 aur -45 pe zrur hoga perfect square.
bhai (n+44)(n+45) + 1 kuch gadbad lag raha hai...
@sujamait said:(n+44)(n+45)+1 = p^2(n+44)(n+45) = (p-1)(p+1)eitherp+1 = (n+44)(n+45)p-1 = 1, p=2so,3 = (n+44)(n+45) but this is nt possible for +tive in Np+1 = n+45p-1 = n+442p = 2n + 892(p-n) = 89which is not possible.... 0 soln
exactly sirjee.. :P
@chillfactor said:Find all positive integers n for which n ˛ + 89n + 2010 is a perfect square.
n=-47 and -42
hence 0 ???
@sujamait said:bhai fact galat kar rahi hai..
exactly matlab 0 solutions he hai na?/ 😛 I didn't do it this way :P
@sujamait said:^^ fact is wrng so avoid it.(n+44)(n+45)+1 = p^2(n+44)(n+45) = (p-1)(p+1)eitherp+1 = (n+44)(n+45)p-1 = 1, p=2so,3 = (n+44)(n+45) but this is nt possible for +tive in Np+1 = n+45p-1 = n+442p = 2n + 892(p-n) = 89which is not possible.... 0 soln
sir main yahi to keh rah hu k n ki +ve values k liye hmein koi soln nai mil rah hai..
and yes i agree the eqn z wrong..it should be (n+44)(n+45) +30...for which we'll get -47 and -42 as solutions..hence no positive integers
@nick_baba said:sir main yahi to keh rah hu k n ki +ve values k liye hmein koi soln nai mil rah hai..and yes i agree the eqn z wrong..it should be (n+44)(n+45) +30...for which we'll get -47 and -42 as solutions..hence no positive integers
han bhai chill sir cnfrm kar denge bus...koi naye Q karte hei ab
@chillfactor said:Find all positive integers n for which n ˛ + 89n + 2010 is a perfect square.
sir ek edit hai mere post me...the eqn shud be (n+44)(n+45) +30...for which we have -47 and -42 as solutions...hence no +ve value of n
@sujamait said:han bhai chill sir cnfrm kar denge bus...koi naye Q karte hei ab
Ans is 0 only..chill sir confirmed by liking my post! hurrayyyy!!! :D
@sujamait said:han bhai chill sir cnfrm kar denge bus...koi naye Q karte hei ab
sure ...bilkul post karo naya ques..
@nick_baba said:sure ...bilkul post karo naya ques..
I dont have OA
Find all pairs of positive integers (n; m) satisfying 3n^2 + 3n + 7 = m^3
Yeah, 0 is correct.
n² + 89n + 2010 = k²
n² + 89n + 2010 - k² = 0
For n to be integer, D should be perfect sq
89² - 8040 + 4k² = a²
(2k - a)(2k + a) = 119 = 1*119 = 7*17
k = 30 and 6
n² + 89n + 2010 = 30²
n = -74, -15
n² + 89n + 2010 = 6²
n = -47, -42
So, given expression can not be a perfect sq for any positive n.
Find the natural number n such that product of all positive divisors of n is 2011^2016
Find all triples (a; b; c) of natural numbers such that lcm(a; b; c) = a + b + c.
@chillfactor said:Yeah, 0 is correct.n² + 89n + 2010 = k²n² + 89n + 2010 - k² = 0For n to be integer, D should be perfect sq89² - 8040 + 4k² = a²(2k - a)(2k + a) = 119 = 1*119 = 7*17k = 30 and 6n² + 89n + 2010 = 30²n = -74, -15n² + 89n + 2010 = 6²n = -47, -42So, given expression can not be a perfect sq for any positive n.Find the natural number n such that product of all positive divisors of n is 2011^2016
sirji..n is +tive integer..so with only D perfect square..se baat ban jayegi kya ? shouldn it be D shall be a perfect square = k and k > 90 and odd :P
@chillfactor said:
Find the natural number n such that product of all positive divisors of n is 2011^2016
6 ke divi ..1,2,3,6 product = 6^2,so formula is N^nooffact/2
N^#/2 = 2011^2016
is it possible ?
@chillfactor said:
Find the natural number n such that product of all positive divisors of n is 2011^2016
No such number ??
Approach:
For any natural number let's say N = x^a*y^b*z^c
product of factors = (x^a*y^b*z^c) ^[(a+1)*(b+1)*(c+1)/2]
Here 2011 is a prime number, so
product of factors = 2011^[(1+1)/2] = 2011 which is not equal to 2011^2016
Hence no such number ..
EDIT: This is wrong approach... please ignore
@sujamait said:I dont have OAFind all pairs of positive integers (n; m) satisfying 3n^2 + 3n + 7 = m^3
It should be 0. Only odd values of 'm' need to be checked and even those odd values show a diverging trend.
@sujamait said:I dont have OAFind all pairs of positive integers (n; m) satisfying 3n^2 + 3n + 7 = m^3
0 pairs hai kya sir?? 
Pta ni 0 se kya pyar h mje
Pta ni 0 se kya pyar h mje