Question

Show that if a cube number is divided by $7$, the remainder is $0,1$ or $6$.

Answer 1

$N$ can be of from $7n$ or $N$ is prime to $7$

Then by using fermats theorem if $N$ is prime to $p$ then $N^{p-1}-1$ is divisible by $p$

$\Rightarrow N^{7-1}-1=N^6-1$ is divisible by $7$

$N^6-1={(N}^3-1\left)\right(N^3+1)$

$\Rightarrow {(N}^3-1)$ and ${(N}^3-1)$ is mutiple of $7n$

$\Rightarrow N^3=7n\pm 1$

When $N$ is of form $7n$ remainder if $0$

When $N^3=7n+1$ remainder is $1$

when $N^3=7n-1=7(n-7)+6$ remainder is $6$

Hence proved