There are 15 possible triples of three positive integers whose product is 16.
According to this question, we must find the quantity of all possible triples such that they have a product of 16. By factor decomposition, we have that 16 is equal to the following product of prime numbers:
[tex]16 = 2^{4}[/tex]
There are the following triples considering order of factors:
(i) [tex]\{16, 1, 1\}[/tex], (ii) [tex]\{1, 16, 1\}[/tex], (iii) [tex]\{1, 1, 16\}[/tex], (iv) [tex]\{2, 8, 1\}[/tex], (v) [tex]\{8, 2, 1\}[/tex], (vi) [tex]\{8, 1, 2\}[/tex], (vii) [tex]\{2, 1, 8\}[/tex], (viii) [tex]\{1, 2, 8\}[/tex], (ix) [tex]\{1, 8, 2\}[/tex], (x) [tex]\{1,4,4\}[/tex], (xi) [tex]\{4, 1, 4\}[/tex], (xii) [tex]\{4,4,1\}[/tex], (xiii) [tex]\{2,2,4\}[/tex], (xiv) [tex]\{2, 4, 2\}[/tex], (xv) [tex]\{4,2,2\}[/tex]
There are 15 possible triples of three positive integers whose product is 16.
We kindly invite to check this question on factor decomposition: https://brainly.com/question/2250220
