Abundant Integer

Quote:

Originally Posted by miz.perfect84

An integer n with the property σ (n) > 2n is sain to be abundant.
Suppose that $n=\color{red}2^r\color{black}p$ , where r ≥ 1 and p is odd prime. Show that for every odd prime p we can find values of r for which n is abundant. List all the abundant numbers of this form which are less than 100.

Cheers

For each $p$ there always exist integers $r$ such that $2^{r+1}>p+1.$ Hence there exist integers $r$ such that

$\sigma(n)\ =\ \sigma(2^rp)$

$=\ \sigma(2^r)\sigma(p)$
$=\ (1+2+2^2+\cdots+2^r)(p+1)$
$=\ (2^{r+1}-1)(p+1)$
$=\ 2^{r+1}p+2^{r+1}-p-1$
$>\ 2^{r+1}p$
$=\ 2n$