The power of $\mu$ and $\sigma$

First, you should know this fact

Fact :

Let $f$ be a multiplicative function and $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{k}^{\alpha_{k}}$ , where $k \in \mathbb{N}$ , be the canonical form of positive integer $n$ .

Then $\sum_{d|n} \mu(d) f(d)=\prod_{i=1}^{k} (1-f(p_{i}))$

And you should know the fact that $\sigma(n)$ is a multiplicative function.

Now, based from fact above, let $n=2017!$ . Then assume $2017!=2^{\alpha_{1}}3^{\alpha_{2}}5^{\alpha_{3}}...p_{k}^{\alpha_{k}}$ , Because $\sigma(n)$ is a multiplicative function, then

$N=\sum_{d|2017!} \mu(d) \sigma(d)=(1-\sigma(2))(1-\sigma(3))(1-\sigma(5))...(1-\sigma(p_{k}))$

Because $1-\sigma(p_{i})=1-(p_{i}+1)=-p_{i}$ , then

$N=(-2)(-3)(-5)...(-p_{k})$ $|N|=2.3.5...p_{k}$

Give a conclusion that the unit digit of $|N|$ is $\boxed{0}$

Let $d$ be an odd divisor of $2017!$ . Then

$\mu(2d) \sigma(2d) = (-1) \mu(d) (2 + \sigma(d)) = -2 \mu(d) - \mu(d) \sigma(d)$

Thus,

$\mu(d) \sigma(d) + \mu(2d) \sigma(2d) = -2 \mu(d)$

By pairing each odd divisor $d$ of $2017!$ to $2d$ , the sum collapses to simply $\sum_{d} -2\mu(d)$ where $d$ is taken over all odd divisors of $2017!$ . (Note that all other divisors $d'$ are divisible by 4, and thus $\mu(d') = 0$ , so they don't contribute to the sum.

Now, what? Well, we continue again. Let $d$ be a divisor of $2017!$ not divisible by either 2 or 3. Then

$\mu(3d) = (-1)\mu(d)$

Thus,

$-2\mu(d) + (-2)\mu(3d) = (-2)(\mu(d) - \mu(d)) = 0$

By pairing each divisor $d$ of $2017!$ that is not divisible by either 2 or 3 to $3d$ , the sum vanishes (their terms cancel). All other divisors $d'$ are divisible by 9 and thus $\mu(d') = 0$ , so they don't contribute to the sum either.

Thus the sum is $\boxed{0}$ .