- Determine the number of unique positive integer solutions.
- Determine the number of unique integer solutions.
Solution for Question 1 Because the product of $x$, $y$ and $z$ is $360$, each of them must be a factor of $360$. Because $360=2^3 \times 3^2 \times 5$, $360$ has 3 prime factors, $2$, $3$ and $5$. Therefore, each of $x$, $y$ and $z$ can be expressed in the form of $$2^a \times 3^b \times 5^c$$ where $a$, $b$ and $c$ are non-negative integers. Let $$ \begin{array}{rcl} x& = & 2^{x_2} \times 3^{x_3} \times 5^{x_5} \\ y& = & 2^{y_2} \times 3^{y_3} \times 5^{y_5} \\ z& = & 2^{z_2} \times 3^{z_3} \times 5^{z_5} \\ \end{array} $$ We have $$ x \times y \times z = 2^{x_2 + y_2 + z_2}\times 3^{x_3 + y_3 + z_3}\times 5^{x_5 + y_5 + z_5} = 2^3 \times 3^2 \times 5^1 $$ Therefore, we have $$ \begin{array}{cccccr} x_2 & + & y_2 & + & z_2 & = & 3 \\ x_3 & + & y_3 & + & z_3 & = & 2 \\ x_5 & + & y_5 & + & z_5 & = & 1 \\ \end{array} $$ The numbers of non-negative integer solutions for the above 3 equations are $$\begin{array}{ccccc} {3 + 3 – 1 \choose 3 – 1} & = & {5 \choose 2} & = & 10 \\ {2 + 3 – 1 \choose 3 – 1} & = & {4 \choose 2} & = & 6 \\ {1 + 3 – 1 \choose 3 – 1} & = & {3 \choose 2} & = & 3 \\ \end{array} $$ So the answer to the question is $ 10\times 6\times 3 = 180 $
Solution for Question 2 Because $x$, $y$ and $z$ can be negative integers, either all of them must be positive, or 2 of them must be negative. So the answer to the question is $$180 + 180 \times {3 \choose 2} = 180 + 180\times 3 = 720$$