Question

Find the number of positive integral solutions of $xyz=30$.
A. 9
B. 27
C. 81
D. 243

Answer 1

Hint: We first need to find the divisors of 30. We have to assign those divisors in the form multiplications of terms x, y, z. We form the multiplications of the divisors and find the number of ways. Then we assign those digits to the unknown terms to find out the solution of the problem.

Complete step-by-step answer:
The equation $xyz=30$ represents the multiplication of three numbers giving value of 30. Those three numbers have to be positive integers. So, $x,y,z\in {{\mathbb{Z}}^{+}}$.
As 30 is made of multiplication of three numbers all three numbers will be divisors of 30. This means 30 is the multiple of all three numbers.
We find out the possible divisors of 30 satisfying the condition $x,y,z\in {{\mathbb{Z}}^{+}}$.
List of the divisors of 30 is 1, 2, 3, 5, 6, 10, 15, 30.
So, there are 8 possible choices for the terms x, y, z with the condition $x,y,z\in {{\mathbb{Z}}^{+}}$.
Now we need to express these numbers into their multiple forms to find value 30.
These multiple forms are $1\times 1\times 30=30$, $1\times 2\times 15=30$, $1\times 3\times 10=30$, $1\times 5\times 6=30$, $2\times 3\times 5=30$.
So, we got 5 such combinations for the terms x, y, z with the condition $x,y,z\in {{\mathbb{Z}}^{+}}$.
But we haven’t assigned the numbers to individual terms.
We will use the theorem of permutation and combination to find the answer.
Out of those 6 possible multiplication combinations, in only 1 case $1\times 1\times 30=30$ we have repetitions.
So, in that case the number of ways the digits 1, 1, 30 can be assigned to the terms x, y, z is $\dfrac{3!}{2!}=3$.
In all the other cases $1\times 2\times 15=30$, $1\times 3\times 10=30$, $1\times 5\times 6=30$, $2\times 3\times 5=30$ the digits are unique. So, the number of ways those combinations can be assigned to the terms is $3!=6$.
Now we find the total number of possible combinations.
They are $3+6\times 4=27$.
So, the number of positive integral solutions of $xyz=30$ is 27.

Note: We need to be careful about when we are assigning the digits to the unknown terms. We need to watch out for the repetitive digits in the combination. We also need to be careful about not repeating the combination of multiplications.