Question

The number of positive integral solutions of $xyz=30$ is:
a) 9
b) 27
c) 81
d) 243

Answer 1

Hint: Here, first find all the factors of 30 and then arrange them in groups so that the product of the numbers will be 30. Then, with the help of permutation, find the number of ways to arrange the numbers in the group. Calculate this for each group. Then, the total number of ways will be taken as the number of positive integral solutions of $xyz=30$.

Complete step-by-step answer:

Here, we have to find the number of positive integral solutions of $xyz=30$.

First we need to find the values of x, y and z whose product will be 30 and then we can find the number of possible arrangements of these numbers.

We know that the factors of 30 are, 1. 2, 3, 5, 6, 10, 15 and 30.

Now, we have to group these factors in such a way that their product is 30.

The groups will be:

30, 1, 1

15, 2, 1

10, 3, 1

6, 5, 1

5, 3, 2

From the above group the values of x, y and z can be interchanged as the multiplication is commutative in real numbers.

We know that the number of ways to arrange n objects = n!

The number of ways to arrange n objects with r objects repeated = $\dfrac{n!}{r!}$

Now, let us find out the number of ways to arrange the group of numbers.

First consider the group 30, 1, 1. Here, there are 3 numbers and 1 is repeated twice. Therefore we can arrange 30, 1, 1 in $\dfrac{3!}{2!}$ ways where:

$\dfrac{3!}{2!}=\dfrac{1\times 2\times 3}{1\times 2}$

Next, by cancellation, we obtain:

$\dfrac{3!}{2!}=3$

So, we can arrange 30, 1, 1 in 3 ways.

Similarly, for the group, 15, 2, 1 there is no repetition of numbers. Hence, I can arrange them in 3! ways where:

$\begin{align}

  & 3!=1\times 2\times 3 \\

 & 3!=6 \\

\end{align}$

Therefore, we can arrange 15, 2, 1 in 6 ways.

Now, consider 10, 3, 1 here also no number is repeated. Therefore, we can arrange them in:
3! = 6 ways.

Similarly, 6, 5, 1 can be arranged in 6 ways.

5, 3, 2 is also arranged in 6 ways.

Therefore, the total number of arrangements = 3 + 6 + 6 + 6 + 6 = 27

Hence, the total number of ways in which the product of $xyz=30$ is 27.

Therefore, we can say that the number of positive integral solutions of $xyz=30$ is 27.

So, the correct answer for this question is option (b).

Note: We can also solve this problem in a short span of time by considering the prime factors of 30 which are 2, 3 and 5. Now, consider x, y and z and 2, 3 and 5 should be kept at any of the three places. Therefore, the total number of integral solutions = $3\times 3\times 3=27$.