Question

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