Question

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

• A. 720
• B. 724
• C. 696
• D. 842

Answer 1

Answer: (A)

We have
$m=\left(9n^2+54n+80\right)\left(9n^2+45n+54\right)\left(9n^2+36n+35\right)$
$=\left[\left(9n^2+30n+24n+80\right)\right]\left[\left(9n^2+27n+18n+54\right)\right]$
$\left[\left(9n^2+15n+21n+35\right)\right]$
$=[3n(3n+10)+8(3n+10)][9n(n+3)+18(n+3)]$
$[3n(3n+5)+7(3n+5)]$
$=(3n+10)(3n+8)(n+3)(9n+18)(3n+5)(3n+7)$
$=(3n+5)(3n+6)(3n+7)(3n+8)(3n+9)(3n+10)$
Now, we know that
$n(n+1)(n+2)\dots (n+r-1)$ is divisible by $r!$
$\therefore m$ is divisible by $6!$ i.e, $720$ .