Originally Posted by
miz.perfect84 An integer n with the property σ (n) > 2n is sain to be abundant.
Suppose that $\displaystyle n=\color{red}2^r\color{black}p$, 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 $\displaystyle p$ there always exist integers $\displaystyle r$ such that $\displaystyle 2^{r+1}>p+1.$ Hence there exist integers $\displaystyle r$ such that
$\displaystyle \sigma(n)\ =\ \sigma(2^rp)$
$\displaystyle =\ \sigma(2^r)\sigma(p)$
$\displaystyle =\ (1+2+2^2+\cdots+2^r)(p+1)$
$\displaystyle =\ (2^{r+1}-1)(p+1)$
$\displaystyle =\ 2^{r+1}p+2^{r+1}-p-1$
$\displaystyle >\ 2^{r+1}p$
$\displaystyle =\ 2n$