Question

The product of three consecutive positive integers is divisible by (a) 4 (b) 6 (c) no common factor (d) only 1

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{Three consecutive positive integers}$

$\underline{\textbf{To find:}}$

$\textsf{The product of three consecutive positive integers is divisible by}$

$\mathsf{(a)4\;(b)6\;(c)\,no\;common\,factor\;(d)\,only\;1}$

$\underline{\textbf{Solution:}}$

$\textsf{We apply Combination formula to solve the problem}$

$\textsf{We know that,}$

$\boxed{\mathsf{n_{C_r}=\dfrac{n!}{r!\;(n-r)!}}}$

$\textsf{Let the three consecutive positive integers be r+1, r+2 and r+3}$

$\mathsf{Consider,}$

$\mathsf{\dfrac{(r+1)\times(r+2)\times(r+3)}{6}}$

$\textsf{This can be written as,}$

$\mathsf{=\dfrac{1\times\,2\times\,3\times\,4\times\,\;.\;.\;.\;.\;.\times\,r\times\,(r+1)\times(r+2)\times(r+3)}{(1\times\,2\times\,3\times\,4\times\,\;.\;.\;.\;.\;.\times\,r)(1\times\,2\times\,3)}}$

$\mathsf{=\dfrac{(r+3)!}{r!\,3!}}$

$\mathsf{=^(r+3)_{C_3}\;it\;is\;an\;integer}$

$\implies\mathsf{\dfrac{(r+1)\times(r+2)\times(r+3)}{6}\;is\;an\;integer}$

$\therefore\mathsf{(r+1)\times(r+2)\times(r+3)\;is\;divisible\;by\;6}$

$\underline{\textbf{Answer:}}$

$\mathsf{Option\;(b)\;is\;correct}$

$\underline{\textbf{Find more:}}$

10C2,11C3,12C4,13C5 find the values​

https://brainly.in/question/38394541

Find r,if 11c4+11c5+12c6+13c7=14cr

https://brainly.in/question/14451378

#SPJ3

Answer 2

Answer:

The product of three

consecutive positive integers is

divisible by

(a) 4

(b) 6

(c) no common factor

(c) only 1