Question

Prove that if a positive number is of the form $6q+5$, then it is of the form $3q+2$ for some integer, but not conversely

Answer 1

Let $n$ be any positive integer.

And the three consecutive positive integers are $n,n+1\mathrm{and}n+2$.

We know that any positive integer can be of the form $6q\mathrm{or}6q+1\mathrm{or}6q+2\mathrm{or}6q+3or6q+4or6q+5$. {From Euclid’s division lemma for $b=6$}

Therefore for $n=6q$,

$\Rightarrow n(n+1)(n+2)=6q(6q+1)(6q+2)$

$=6\left[q\right(6q+1\left)\right(6q+2\left)\right]$

$=6m$, which is divisible by $6$……..$[m=q(6q+1\left)\right(6q+2\left)\right]$

For $n=6q+1$,

$\Rightarrow n(n+1)(n+2)=(6q+1)(6q+2)(6q+3)$

$=6\left[\right(6q+1\left)\right(3q+1\left)\right(2q+1\left)\right]$

$=6m$, which is divisible by $6$……… $[m=(6q+1\left)\right(3q+1\left)\right(2q+1\left)\right]$

For $n=6q+2$,

$\Rightarrow n(n+1)(n+2)=(6q+2)(6q+3)(6q+4)$

$=6\left[\right(3q+1\left)\right(2q+1\left)\right(6q+4\left)\right]$

$=6m$, which is divisible by $6$……….. $[m=(3q+1\left)\right(2q+1\left)\right(6q+4\left)\right]$

For $n=6q+3$,

$\Rightarrow n(n+1)(n+2)=(6q+3)(6q+4)(6q+5)$

$=6\left[\right(2q+1\left)\right(3q+2\left)\right(6q+5\left)\right]$

$=6m$, which is divisible by $6$…………$[m=(2q+1\left)\right(3q+2\left)\right(6q+5\left)\right]$

For$n=6q+4$,

$\Rightarrow n(n+1)(n+2)=(6q+4)(6q+5)(6q+6)$

$=6\left[\right(3q+2\left)\right(3q+1\left)\right(2q+2\left)\right]$

$=6m$, which is divisible by $6$……………$[m=(3q+2\left)\right(3q+1\left)\right(2q+2\left)\right]$

For$n=6q+5$,

$\Rightarrow n(n+1)(n+2)=(6q+5)(6q+6)(6q+7)$

$=6\left[\right(6q+5\left)\right(q+1\left)\right(6q+7\left)\right]$

$=6m$, which is divisible by $6$……………….$[m=(6q+5\left)\right(q+1\left)\right(6q+7\left)\right]$

Hence, the product of three consecutive positive integers is divisible by $6$.