Table of Contents

kPerfect, Abundant, and Deficient Numbers Examples 1
Recall from the kPerfect, Abundant, and Deficient Numbers page the following definitions:
 $n \in \mathbb{N}$ is a perfect number if $\sigma(n) = 2n$, and is a $k$perfect number if $\sigma (n) = kn$ for some $k \in \mathbb{N}$.
 $n \in \mathbb{N}$ is an abundant number if $\sigma(n) > 2n$.
 $n \in \mathbb{N}$ is a deficient number if $\sigma(n) < 2n$.
We will now look at some problems involving these type of numbers.
Example 1
Suppose that $n \in \mathbf{N}$ is an odd $4$perfect number. Determine what value of $k$ makes $4n$ a $k$perfect number.
We first note that since n is odd, $2$ does not appear in the prime power decomposition of of $n$. Consider the prime power decomposition of $n$:
(1)Then the prime power decomposition of $4n$ is:
(2)Hence we have that:
(3)Hence if $n$ is an odd positive integer and is $4$perfect, then $4n$ is a $28$perfect number.
Example 2
If $n$ is a $3$perfect number and $(n, 12) = 1$, then, determine if $12n$ is a $k$perfect number.
We note that since $(n, 12) = 1$, it follows that $\sigma (12n) = \sigma(12) \sigma (n) = \sigma (4) \sigma (3) \sigma (n) = 7 \cdot 4 \sigma (n)$.
But we note that since n is a 3perfect number, then $\sigma (n) = 3n$. Hence:
(4)Thus $12n$ is an $84$perfect number.
Example 3
If $p$ is prime, $n$ is a $p$perfect number, and $(n, p) = 1$, then, determine if $pn$ is a $k$perfect number.
Since $(n, p) = 1$, it follows that $\sigma (pn) = \sigma (p) \sigma (n) = (p + 1) \sigma (n)$.
Since $n$ is a $p$perfect number, it follows that $\sigma(n) = pn$. Hence:
(5)So $pn$ is a $p^2 + p$perfect number.