The number of (a,b,c), where a,b,c are positive integers such that abc = 30, is
27
Finding the Number of Positive Integer Solutions for abc = 30
The problem asks for the number of ordered triples (a, b, c) where a, b, and c are positive integers that satisfy the equation abc = 30.
To solve this, we first need to find the prime factorization of the number 30.
The prime factorization of 30 is:
Here, the prime factors are 2, 3, and 5, and each appears with an exponent of 1. We need to distribute these prime factors among the three positive integers a, b, and c such that their product is 30. Since a, b, and c must be positive integers, each of a, b, and c must be a product of some subset of these prime factors. If a variable doesn't receive a particular prime factor, it implicitly has that prime factor raised to the power of 0 (which is 1).
Consider the prime factor 2. The factor '2' must go into either a, b, or c. It cannot be split, and it must be assigned somewhere. There are 3 distinct choices for where the factor '2' goes: a, b, or c.
Consider the prime factor 3. Similarly, the factor '3' must go into either a, b, or c. There are 3 distinct choices for where the factor '3' goes: a, b, or c.
Consider the prime factor 5. The factor '5' must go into either a, b, or c. There are 3 distinct choices for where the factor '5' goes: a, b, or c.
Since the decisions for each prime factor are independent, the total number of ways to distribute the prime factors 2, 3, and 5 among the three variables a, b, and c is the product of the number of choices for each prime factor.
Total number of solutions = (Number of choices for factor 2) \(\times\) (Number of choices for factor 3) \(\times\) (Number of choices for factor 5)
Total number of solutions = \(3 \times 3 \times 3 = 3^3\)
Calculating the result:
Each unique distribution of these prime factors results in a unique ordered triple (a, b, c) of positive integers whose product is 30. For example:
- If 2 goes to a, 3 goes to b, 5 goes to c, the triple is (2, 3, 5).
- If 2 goes to a, 3 goes to c, 5 goes to b, the triple is (2, 5, 3).
- If 2, 3, and 5 all go to a, the triple is (30, 1, 1).
- If 2 and 3 go to a, 5 goes to b, and c gets none (which means c=1), the triple is (6, 5, 1).
The number of ways to distribute \(m\) distinct items into \(k\) distinct bins is \(k^m\). In this case, we are distributing 3 distinct prime factors (2, 3, 5) into 3 distinct bins (a, b, c).
Number of solutions = \(3^3 = 27\).
Therefore, there are 27 different ordered triples (a, b, c) of positive integers such that abc = 30.
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Prime Factorization | Breaking down a number into its prime factors. For 30, it's \(2 \times 3 \times 5\). |
| Positive Integers | The set of numbers {1, 2, 3, ...}. Includes 1. |
| Distributing Factors | Assigning each prime factor (or its power) to one of the variables (a, b, c). |
| Multiplication Principle | If there are \(n_1\) ways to do one thing and \(n_2\) ways to do another independent thing, there are \(n_1 \times n_2\) ways to do both. Used here for distributing each distinct prime factor. |
Additional Information: Generalizing the Problem
Consider a more general case: finding the number of positive integer solutions to \(x_1 x_2 \dots x_k = N\), where \(N = p_1^{e_1} p_2^{e_2} \dots p_m^{e_m}\) is the prime factorization of \(N\).
For each prime factor \(p_i\) raised to the power \(e_i\), we need to distribute the \(e_i\) occurrences of \(p_i\) among the \(k\) variables \(x_1, \dots, x_k\). Let \(x_j = p_1^{a_{j1}} p_2^{a_{j2}} \dots p_m^{a_{jm}}\). The condition \(x_1 x_2 \dots x_k = N\) means that for each prime \(p_i\), the exponents must sum up:
where \(a_{ji}\) are non-negative integers (since \(x_j\) must be positive integers, the exponents must be non-negative). The number of non-negative integer solutions to this equation is given by the stars and bars formula: \(\binom{e_i + k - 1}{k - 1}\).
Since the distribution of each prime factor \(p_i\) is independent, the total number of positive integer solutions for \(x_1 x_2 \dots x_k = N\) is the product of the number of ways to distribute each prime factor:
In our specific problem, \(N=30\), \(k=3\). The prime factorization is \(30 = 2^1 \times 3^1 \times 5^1\). So, \(p_1=2, e_1=1\); \(p_2=3, e_2=1\); \(p_3=5, e_3=1\); and \(m=3\).
Number of solutions = \(\binom{1 + 3 - 1}{3 - 1} \times \binom{1 + 3 - 1}{3 - 1} \times \binom{1 + 3 - 1}{3 - 1}\)
Number of solutions = \(\binom{3}{2} \times \binom{3}{2} \times \binom{3}{2}\)
Since \(\binom{n}{k} = \binom{n}{n-k}\), \(\binom{3}{2} = \binom{3}{3-2} = \binom{3}{1} = 3\).
Number of solutions = \(3 \times 3 \times 3 = 27\).
This confirms the result obtained by considering each distinct prime factor separately and multiplying the choices.
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