k-Perfect, Abundant, and Deficient Numbers Examples 1

k-Perfect, Abundant, and Deficient Numbers Examples 1

Recall from the k-Perfect, 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)
\begin{align} \quad n = p_1^{e_1}p_2^{e_2}...p_k^{e_k} \end{align}

Then the prime power decomposition of $4n$ is:

(2)
$$4n = 2^2 p_1^{e_1}p_2^{e_2}...p_k^{e_k}$$

Hence we have that:

(3)
\begin{align} \sigma(4n) & = \sigma(2^2)\sigma (p_1^{e_1}p_2^{e_2}...p_k^{e_k}) \\ & = 7\sigma (n) \\ & = 7(4n) \\ \end{align}

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 3-perfect number, then $\sigma (n) = 3n$. Hence:

(4)
\begin{align} \quad \sigma (12n) &= 7 \cdot 4 \cdot (3n) \\ \quad &= 84n \end{align}

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)
\begin{align} \quad \sigma (pn) &= \sigma(p) \sigma(n) \\ \quad &=(p + 1)pn \\ \quad &= (p^2 + p)n \end{align}

So $pn$ is a $p^2 + p$-perfect number.