Can an odd perfect number be divisible by $5313$? I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $5313$.
 A: It is quite easy to prove that an odd perfect prime has a factorisation into powers of odd primes, where exactly one power is a prime of the form 4k+1, raised to a power 4n+1 (n can be 0), and all the other powers are even. This implies an odd perfect number which is a multiple of $5313 = 3 * 7 * 11 * 23$ must be a multiple of $3^27^211^223^2$. $3^2 7^2 11^2 23^2$ has an abundancy of 1.9037 < 2, so multiples of this number could a priori have an abundance of 2. 
However, since $1 + 3 + 9 = 13$, if the number is not a multiple of $3^4$ then it must be a multiple of 13 (it doesn't have to be a multiple of $13^2$ because 13 = 4k+1). Since 1 + 7 + 49 = 57, if the number is not a multiple of $7^4$ then it must be a multiple of 3 (which it is) and of 19, and therefore a multiple of 19^2. 
Each of the four possibilities $3^4 7^4 11^2 23^2$, $3^4 7^2 11^2 23^2 19^2$, $3^2 7^4 11^2 23^2 13$ and $3^2 7^2 11^2 23^2 19^2 13$ has an abundancy just above 2, so these numbers and any multiples of them cannot be perfect numbers.
A: As $\textbf{Ron Ford}$ already pointed out in the comments, an odd perfect number can be divisible by $5313 = 3\cdot7\cdot11\cdot23$ only if it is divisible by $3^47^411^423$. To see this, use the same line of reasoning as in the linked question.
Continue by noting that
$$\left(\sum_{k=0}^4 \frac{1}{3^k}\right)\left(\sum_{k=0}^4 \frac{1}{7^k}\right)\left(\sum_{k=0}^4 \frac{1}{11^k}\right)\left(1 + \frac{1}{23}\right) > 2,$$
which implies that the abundancy index of the number must be larger than $2$. This contradicts the definition of a perfect number.
