Question

product of any four consecutive positive integers is divisible by

Answer 1

$\textbf{To prove:}$

$\text{Product of any four consecutive positive integers is divisible by 24}$

$\textbf{Solution:}$

$\text{Let the four consecutive integers be n,n+1,n+2 and n+3}$

$\text{Consider,}$

$\dfrac{n(n+1)(n+2)(n+3)}{24}$

$=\dfrac{n(n+1)(n+2)(n+3)}{1{\times}2{\times}3{\times}4}$

$=\dfrac{n(n+1)(n+2)(n+3)}{4!}$

$\text{Multiply both numerator and denominator by (n-1)! }$

$=\dfrac{(n-1)!n(n+1)(n+2)(n+3)}{(n-1)!\,4!}$

$=\dfrac{(n+3)!}{(n-1)!\,4!}$

$=\dfrac{(n+3)!}{((n+3)-4)!\,4!}$

$\text{We know that,}$

$\boxed{\bf\,n_C_r=\dfrac{n!}{(n-r)!\,r!}}$

$=(n+3)_C_4\;\text{is a integer}$

$\implies\dfrac{n(n+1)(n+2)(n+3)}{24}=\text{an integer}$

$\therefore\text{The product}\bf\,n(n+1)(n+2)(n+3)\;\textbf{is divisible by 24}$

$\textbf{Hence proved}$