# Suppose a number n is k-perfect and its prime divisors are single digit numbers. Prove k < 5.

Let $\sigma(n)$ be the sum of the positive divisors of an integer n. A number is k-perfect if $\sigma(n)$ = kn.

Let n be k-perfect. I know that $n=2^a\cdot3^b\cdot5^c\cdot7^d$ where a, b, c, d $\geq0$. Since each of the factors of n are coprime, then I know that $\sigma(n)=\sigma(2^a)\sigma(3^b)\sigma(5^c)\sigma(7^d)$. Since n is k-perfect, then

$k\cdot2^a\cdot3^b\cdot5^c\cdot7^d=\sigma(2^a)\sigma(3^b)\sigma(5^c)\sigma(7^d)$

$k = \frac{\sigma(2^a)\sigma(3^b)\sigma(5^c)\sigma(7^d)}{2^a\cdot3^b\cdot5^c\cdot7^d}$

I have previously proven that for a prime p, and exponent e, then $\frac{\sigma(p^e)}{p^e}<p$. Therefore, I can find that $k<2\cdot3\cdot5\cdot7$. However, I'm not sure where to go from here to further reduce this to $k<5$. Any suggestions are appreciated.

$$\frac{\sigma(p^c)}{p^c} = 1 + p^{-1} + \ldots p^{-c} < \sum_{j=0}^\infty p^{-j} = \frac{1}{1-1/p}$$
For what it's worth, there is also the lower bound $$\frac{\sigma(p^c)}{p^c} = 1 + \ldots + p^{-c} \geq 1 + p^{-1} = \frac{\sigma(p)}{p} = \frac{p+1}{p} > 1,$$ so that $$k \geq \frac{\sigma(2)}{2}\cdot\frac{\sigma(3)}{3}\cdot\frac{\sigma(5)}{5}\cdot\frac{\sigma(7)}{7} = \frac{3}{2}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{8}{7} = \frac{96}{35} > 2 \implies k \geq 3.$$