Probability of a divisor being odd

Recall that a number $N$ with prime factorization

$p_1^{q_1}p_2^{q_2}p_3^{q_3}p_4^{q_4} \cdots p_k^{q_k}$

has $\left(q_1 + 1\right)\left(q_2 + 1\right)\left(q_3 + 1\right)\left(q_4 + 1\right) \cdots \left(q_k + 1\right)$ factors .

Now,

$15! = 2^{11} \times 3^6 \times 5^3 \times 7^2 \times 11 \times 13$

has $(11 + 1)(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1)$ factors.

For the product of two numbers to be odd, they must both be odd, so the odd factors of $15!$ must be all possible combinations of products of its odd prime factors - namely, $3$ , $5$ , $7$ , $11$ , and $13$ . Thus, $15!$ has $(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1)$ odd factors, making the probability that a randomly selected factor is odd

$\dfrac{(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1)}{(11 + 1)(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1)} = \boxed{\dfrac{1}{12}}$

The prime factorization of $1307674368000$ is $2^{11} \cdot 3^6 \cdot 5^3 \cdot 7^2 \cdot 11^1 \cdot 13^1$ .

The probability of a divisor being odd is $\frac{(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1)}{(11 + 1)(6 + 1)(3 + 1)(2 + 1)(1 + 1)(1 + 1)}\ = \frac{1}{12}$