Question

The product of five consecutive natural numbers is divisible by

• A. 10
• B. 20
• C. 30
• D. 120

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A 10
B 20
C 30
D 120

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

Here, given $k=5$, so $k!=120$ which is divisible by $10,20,30,120$