Question

The greatest positive integer. which divides $(n+16)(n+17)(n+18)(n+19)$, for all $nϵN$, is

• A. 2
• B. 4
• C. 24
• D. 120

Answer 1

The correct option is C 24

Let $k$ consecutive natural number be $(n+1),(n+2)........,(n+k)$

Thus their product is, $P=(n+1)(n+2)(n+3).........(n+k)$

$\Rightarrow P=\frac{n!(n+1)(n+2)........(n+k)}{n!}=\frac{(n+k)!}{n!k!}\times k!{=}^{n+k}{C}_k\times k!=k!\times$ Integer

Hence product of $k$ natural number is always divisible by $k!$

Here, $n$ is replaced by $n+15$ and $k=4$

Therefore $(n+16)(n+17)(n+18)(n+19)$ is divisible by $k!=4!=24$