Question

let m=(9m^2+54n+80)(9n^2+45n+54)(9n2+36n+35). the greatest positive integer which divides m for all positive integers n is​

Answer 1

Let f(m,n) = 45m + 36n , where m and n are integers (positive or negative) what is the minimum positive value for f(m,n) for all values of m,n.

The greatest common divisor of two non zero positive integers 'a' and 'p' Is the smallest positive integer that can be written as “ax+by".

Therefore, the smallest positive value say 'p' that can be written as 45m+36n is the greatest common divisor of 45 and 36.

Consider factors of 45 and 36, 45=3x3x5

36 =2x2x3x3

p = GCD of 45 and 36 =3 x 3 p=9

Therefore, the minimum positive value for f(m,n) for all values of m,n is 9.

hope it helps you....

Answer 2

concept

the concept used in this question is the greatest positive integer

given

m=(9m^2+54n+80)(9n^2+45n+54)(9n2+36n+35).

find

to find the greatest positive integer which divides m for all positive integers n

solution

m= [(9n²+30n+24n+80)] [(9n²+27n+18n+54) [(9n²+15n+21n+35)]

=[3n(3n+10)+8(3n+10)] [9n(n+3)+18(n+3)] [(3n(3n+5+7(3n+5)]

=(3n+10)(3n+8)(n+3)(9n+18)(3n+5)(3n+7)

=(3n+5) (3n+6) (3n+7) (3n+8) (3n+9) (3n+10)

now, we know that

n(n+1) (n+2).....(n+r-1) is divisible by r

∴m is divisible by 6 i.e 720

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