Problem of 2017(II)

$2017$ is a prime number. So it's factors are $\pm1$ and $\pm2017$

So it can be possible for three positive number's multiplication or two negative and one positive number's multiplication.

$\color{#D61F06}1$ Thus the ans could be $1,1,2017$ . Arranging them, here we get for three variable $\frac{3!}{2!}=3$ solution.

$\color{#20A900}2$ Another could be $-1,-1,2017$ . Arranging them, here we get for three variable $\frac{3!}{2!}=3$ solution.

$\color{#3D99F6}3$ Another could be $1,-1,-2017$ . Arranging them, here we get for three variable $3!=6$ solution.

Now total solution is $3+3+6=\boxed{\color{#3D99F6}12}$

The solution are:

$\begin{matrix} x & & y & & z \\ 1 & & 1 & & 2017 \\ 1 & & 2017 & & 1 \\ 2017 & & 1 & & 1 \\ -1 & & -1 & & 2017 \\ -1 & & 2017 & & -1 \\ 2017 & & -1 & & -1 \\ -1 & & 1 & & -2017 \\ -1 & & -2017 & & 1 \\ 1 & & -1 & & -2017 \\ 1 & & -2017 & & -1 \\ -2017 & & 1 & & -1 \\ -2017 & & -1 & & 1 \end{matrix}$

Here 1st three are for No. $\color{#D61F06}1$ , After three are for No. $\color{#20A900}2$ and last six are for No. $\color{#3D99F6}3$ .

Since 2017 is a prime, the solutions of $(x,y,z)$ can only be combinations of $(1,1,2017)$ if all three of them are positive. There are also solutions when two of $x$ , $y$ and $z$ are negative.

• The number of solutions for all positive $x$ , $y$ and $z$ is arranging two integers in three places $N_{+++} = \dbinom 32 = 3$ .
• For two negative $x$ , $y$ and $z$ , the number of ways to arrange $|x|$ , $|y|$ and $|z|$ is $N_{| \cdot|} = = \dbinom 32 = 3$ , and for each combination the number of ways to assign 2 negative signs to three numbers $N_{2-} = = \dbinom 32 = 3$ , together $N_{+--} = N_{| \cdot|} \times N_{2-} = 9$

Therefore, the total number of solutions of $(x,y,z)$ , $N = N_{+++}+N_{+--} = 3+9 = \boxed{12}$ .

Since $2017$ is a prime number, the only 2 divisors of $2017$ is $1$ and $2017$ .

Thus, we have three un-ordered tripulets of solution: $(1, 1, 2017), (-1, -1, 2017), (-1, 1, -2017)$ .

We should match these numbers with $x, y, z$ , and the number of the ways of distribution is $\large\frac{3!}{2}$ in the first 2 cases repectively, and $3!$ in the last case.

Therefore, the answer is $3 \times 2+6=\boxed{12}$