x=7^ay=7^bz=7^ca+b+c = 11total = 13c2 = 78but when 2 are same -> (0,0,11);(1,1,9);(2,2,7);(3,3,5);(4,4,3);(5,5,1) ->6*3 = 18 => 78-18 = 6060/6 = 10so total cases = 10+6 = 16...
and if i have to find triplets where no two are same cant i do it by distributing 0,1,2 to a,b,c respectively and then finding the number of positive integral solutions? if it can be done kindly show the method as i dont know the formula to find the number of positive integral solutions
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sir ji last baar likh raha hoon 3 values ( 0 , 3 , -1)?????
The two equations can be written as (y-m)^2 = 5m^2 + 4, and (y-2)^2 = 2(m^3 + m +2)
For three different roots either one of the equations should have a double root while the other should not or both the equations should have a common root.
(y-m)^2 = 5m^2 + 4 cannot have double roots as RHS can never be zero.
If the second equation has double root, then (m+1)(m^2 - m + 2) = 0 => m = -1
But if we put m = -1 in the two original equations we get the 2, -4 as roots for the first equation and 2 as double root for the second. Thus, in all we have 2 different roots. Hence, m = -1 is also ruled out.
Also, if both the equations have a common root, then (y^2-2my-4(m^2 + 1)) - (y^2 €“ 4y -2m(m^2 + 1)) = 0 hence, (2m - 4)y = (2m - 4)(m^2 + 1)=> m =2 or y=m^2 + 1
Putting m = 2 in our two quadratic equations we get (m-2)^2 = 24 for both=> we have only 2 different roots.
Putting y = m^2 + 1, we get (m^2 - 1)^2 = 2(m^3 + m + 2) which can be factored as (m+1)(m-3)(m^2 + 1) = 0, hence m = -1 or 3.
But we already neglected m = -1. So for m=3 we get the equations as y^2-6y-40 = 0, and y^2-4y-60 = 0. The roots of first eq. are 10, -4 and that of second are 10, -6.
So only for m = 3 there are three different roots 10,-4, and -6
Consider with 9 and 4.All terms divisible by 4. So of the form 4kAlso when divided by 9, the terms leave -1, 0, 0, -1, 0, 0....which in 99 terms adds to -33 equiv +3 with 9. So 9m+3 form as well.1st number satisfying both = 12.regardsscrabbler
bro isko summation wala logic laga ke bhi toh solve kar sakte hai.. you will get (summation)n^2*(n+1)^3 = [n*(n+1)*(2n+1)]/6*[((n+1)*(n+2))/2]^2