Question

Show that every number and its cube when divided by $6$ leave the same remainder.

Answer 1

Difference of the cube and its number $=n^3-n$

$=n(n^2-1)=n(n-1)(n+1)=(n-1)n(n+1)$

When ever a number is divided by $3$ it leaves a remainder $0,1$ or $2$

So the number is of form $n=3p,3p+1,3p+2$

If $n=3p$ then $n$ is divisible by $3$

If $n=3p+1$ then $n-1=3p+1-1=3p$ which is divisible by $3$

If $n=3p+2$ then $n+1=3p+2+1=3p+3=3(p+1)$ which is divisible by $3$

So of of the consecutive terms is always divisible by $3$

Atleast one of the three consecutive numbers will always be even so it will be divisible by $2$

Hence proved that $(n-1)n(n+1)$ will always be divisible by $6$

So when $n^3-n$ is divided by $6$ it leaves remainder $0$

$\Rightarrow n^3$ and $n$ when divided by $6$ leaves same remainder