Problem 1 · Integer triples with |x|·|y|·|z| = 6
Counting ordered divisor triples and sign choices.
Integer answer, at most 4 digitsHow many triples $(x, y, z)$ of solutions, with $x$, $y$, $z$ integers, does the equation $|x|\cdot|y|\cdot|z| = 6$ have? (Note: $|x|$ denotes the absolute value.)
Reasoned solution
Key idea: split the problem into two independent parts: which absolute values, and which signs.
First count ordered triples of positive integers $(|x|,|y|,|z|)$ with product $6 = 2\cdot 3$. Each prime factor ($2$ and $3$) must land in exactly one of the three positions: $3 \cdot 3 = 9$ triples. (These are the permutations of $(1,1,6)$ and $(1,2,3)$: $3 + 6 = 9$.)
Now the signs: no value can be $0$, so each of the three variables can independently be positive or negative: $2^{3} = 8$ sign combinations for each triple of absolute values.