Let m=(9n^2+54n+80)(9n^2+45n+54) (9n^2+36n+35). The greatest positive integer which divides m, for all positive integers n, is

Question

Accepted Answer

Correct answer: A. (a) Given, m=(9n^2+54n+80)(9n^2+45n+54)= ((9n^2+30n+24n+80) ) ((9n^2+27n+18n+54) ) ((9n^2+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 ) Now, we know that n (n+1 ) (n+2 ) … (n+r-1 ) is divisible by r ! ∴m is divisible by 6 ! i.e., 720 .

$Let m = (9 n^{2} + 54 n + 80) (9 n^{2} + 45 n + 54) (9 n^{2} + 36 n + 35) . The greatest positive integer which divides m, for all positive integers n, is$

Solution

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