Question

The product of $r$ consecutive integers is divisible by $r!$.

• A. True
• B. False

Answer 1

The correct option is A True

The product of some r consecutive integers can be represented as

$\stackrel{rconsectiveintegers}{\stackrel{}{(n+r)(n+r-1)\cdots (n+1)}}=\frac{(n+r)!}{n!}$

where n is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in $(n+r)!$ appear just as frequently or more as in n!r!, then you are saying that for some integer $k$ (likely big) that $(n+r)!=k\cdot n!r!.$ So your product of n consecutive integers is

$\frac{(n+r)!}{n!}=\frac{k\cdot n!r!}{n!}=k\cdot r!$

and is therefore divisible by $r!.$