Is there an odd integer $x < 105$ for which it is known that $x \nmid N$, if $N$ is an odd perfect number? Is there an odd integer $x < 105$ for which it is known that $x \nmid N$, if $N$ is an odd perfect number?
I have asked the same question in MSE, but did not get any answers.  I was wondering if anybody here has anything to share regarding this.
Thank you!
 A: No, such a result would be a major breakthrough regarding our knowledge on odd perfect numbers. 
A few years ago there was some confusion, since due to careless reading and citing of the article "Every odd perfect number has a prime factor which exceeds $10^6$" by Cohen and Hagis the impression arose that they had proved just such a theorem (which they never claimed).
A: This is too long to be posted as a comment, as I just wanted to share some of my recent thoughts on why it is difficult to obtain such an integer $x \nmid N$, where $N = {q^k}{n^2}$ is a (hypothetical) odd perfect number.
Since a multiple of an abundant number is also abundant, and since the smallest odd abundant number is $945$, we know that
$${3^3}\cdot{5}\cdot{7} = 945 \nmid N.$$
In fact, we know that
$$3\cdot5\cdot7 = 105 \nmid N,$$
although $105$ is deficient.
Note that
$$\{\{3 \mid N\} \land \{5 \mid N\} \land \{7 \mid N\}\} \implies \{105 \mid N\}.$$
By the contrapositive,
$$\{105 \nmid N\} \implies \{\{3 \nmid N\} \lor \{5 \nmid N\} \lor \{7 \nmid N\}\}.$$
However, all existing computational projects which implement factor chains (see this MSE post for a very down-to-earth sample) to check whether $3 \mid N$ or otherwise have not terminated (i.e., AKA "resulted to a contradiction").  Likewise, it is currently unknown whether the smallest possible Euler prime $q = 5$ does divide an odd perfect number.  (It is known though, by work of Iannucci, that $q = 5$ implies $k = \nu_{q}(N) = 1$. Iannucci proved that $q = 5$ implies $k = \nu_{q}(N) = 1$ under some additional assumptions.)
