Question

The number of positive integral solutions of $abc=30$ is?

Answer 1

Hint: In this question we need to find the number of positive integral solutions. So let us consider \[{\text{a b c}}\] as a distinct digit. By finding the factors of \[{\text{30}}\], we can find the digits and the number of ways they can be arranged. From that we can easily find out the number of solutions for the question given here.

Complete step by step answer:
We are given the number of positive integral solutions \[{\text{a b c}} = 30\]
Let us consider that \[{\text{a b c}}\] as a distinct digit.
Factors of \[{\text{30}}\] are \[{\text{1,2,3,5,6,10,15,30}}\].
Among them we can choose the prime factors.
A number which has exactly two factors, number \[{\text{1}}\] and the number itself is called a prime number. According to this, the prime number which has turned into the factors of \[{\text{30}}\] is \[{\text{2,3,5}}\].
\[{\text{a b c}} \Rightarrow \] is a three digit number and we also found three prime factors \[{\text{2,3,5}}\].
Therefore \[{\text{2}}\] can be assigned as \[{\text{a}}\],\[{\text{b}}\] or \[{\text{c}}\], thus we have three options. Similarly \[{\text{3}}\] and \[{\text{5}}\] can also be assigned as \[{\text{a}}\],\[{\text{b}}\] or \[{\text{c}}\] and we have three options for each \[{\text{3}}\] and \[{\text{5}}\].
Hence the number of possible solutions \[{\text{ = 3}} \times {\text{3}} \times {\text{3 = 27}}\].
Therefore the number of positive integral solutions of \[{\text{a b c}} = 30\] is \[27\].

Note:
Alternative Method:
There is also another method to solve this same question without choosing the prime factors. Let us write the possibility of getting \[{\text{30}}\] by multiplying three numbers as we have \[{\text{a b c}}\] here.
\[2 \times 3 \times 5 = 30\], here the possibility of \[{\text{2}}\] is three that is can be as \[{\text{a}}\],\[{\text{b}}\] or \[{\text{c}}\]. This can be say as \[3!\].
Similarly, \[3 \times 10 \times 1 = 30 \Rightarrow 3!\]
\[5 \times 6 \times 1 = 30 \Rightarrow 3!\]
\[15 \times 2 \times 1 = 30 \Rightarrow 3!\]
\[30 \times 1 \times 1 = 30\], and here the possibility of \[30\] is three and it can be as \[{\text{a}}\],\[{\text{b}}\] or \[{\text{c}}\] and the possibility of \[1\] can be both of the digits at a time, this can be written as \[\dfrac{{3!}}{{2!}}\].
Thus the number of solutions \[ = 3! + 3! + 3! + 3! + \dfrac{{3!}}{{2!}}\]
This can be shortly written as:
\[ = (3! \times 4) + \dfrac{{3!}}{{2!}}\]
Expanding the factorials,
\[ = ((3 \times 2 \times 1) \times 4) + \dfrac{{3 \times 2 \times 1}}{{2 \times 1}}\]
Performing multiplication,
\[ = 6 \times 4 + \dfrac{6}{2}\]
Now multiplying \[6 \times 4\] we will get \[24\] and dividing \[\dfrac{6}{2}\] we will get 3,
\[ = 24 + 3\]
Finally adding the numbers we will get,
\[ = 27\]
Therefore the number of positive integral solutions of \[{\text{a b c}} = 30\] is \[27\].