Question

Shew that the highest power of $n$ which is contained in $n^r-1$ is equal to $\frac{n^r-nr+r-1}{n-1}$.

Answer 1

The highest power required is equal to the sum of the integral parts of

$\frac{n^r-1}{n},\frac{n^r-1}{n^2},.....,\frac{n^r-1}{n^{r-1}}$

$=(n^{r-1}-1)+(n^{r-2}-1)+.....+(n-1)$

$=\frac{n^r-n}{n-1}-(r-1)=\frac{n^r-nr+r-1}{n-1}$