Problem 1 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(y\) be an element of the set \(A=\\{1,2,3,5,6,10,15\), 30\\} and \(x_{1}, x_{2}, x_{3}\) be integers such that \(x_{1} x_{2} x_{3}=y\), then the number of positive integral solutions of \(x_{1} x_{2} x_{3}=y\) is (A) 64 (B) 27 (C) 81 (D) None of these

Step-by-Step Solution

Answer

The number of positive integral solutions is 27.

Step 1: Analyze the given problem

We need to find the number of positive integral solutions to the equation \(x_1 x_2 x_3 = y\) for each \(y\) in set \(A = \{1, 2, 3, 5, 6, 10, 15, 30\}\).

Step 2: Find total solutions for one value of y

For each integer \(y\), recall that the number of positive solutions for \(x_1 x_2 x_3 = y\) is equivalent to the number of ways the prime factorization of \(y\) can be distributed among three variables, considering order.For example, if \(y = 30 = 2^1 \cdot 3^1 \cdot 5^1\), the number of positive integral solutions can be found by considering how each prime power can be spread among the three variables: For integer \(a, b, c\), \(a + b + c = 1\) for prime 2\(a + b + c = 1\) for prime 3\(a + b + c = 1\) for prime 5.

Step 3: Apply the combinatorial formula

The formula for counting the number of solutions to each equation \(a + b + c = n\) is given by \(\binom{n + 2}{2}\). Thus, for each prime factor, the number of solutions is \(\binom{1 + 2}{2} = 3\) since \(n=1\).Therefore, the solution for each prime can be represented as: \(3 \times 3 \times 3 = 27\).

Step 4: Verify calculations for different values of y

We need to check whether a similar method gives the same number of solutions for all values in \(A\). For that, we see based on the prime power distribution logic from Step 3, each \(y\) can be similarly broken down irrespective of individual prime variations, meaning each leads to \(3^{n/1} = 27\) solutions (as all are \(1^n \) exponential factors), given the primes facts about \(n = 1\).

Step 5: Conclusion

Since each value of \(y\) leads to the same number of solutions due to the consistent power of 1 in all prime factorizations, we conclude that the total number of solutions for each \(y\) is the same as above.Thus, the number of positive integral solutions for each \(y\) is \(27\), which is option B.

Key Concepts

Positive Integral Solutions

In combinatorics, finding the number of positive integral solutions for an equation means determining how many different sets of positive integers satisfy the equation. For instance, given the equation \( x_1 x_2 x_3 = y \), our task is to find different combinations of \( x_1, x_2, \) and \( x_3 \) that multiply to \( y \), where \( y \) is an element of a specific set. All values for \( x_1, x_2, \) and \( x_3 \) must be positive integers.
To solve for these, one helpful strategy is to use the prime factorization of \( y \). This involves breaking down \( y \) into its prime components and figuring out how to distribute these components among the three variables, \( x_1, x_2, \) and \( x_3 \). Each unique distribution corresponds to a different solution.
This type of problem often involves combinatorial methods, particularly when assessing multiple variables. It's a fascinating process because it combines number theory with combinatorics to find all possibilities.

Prime Factorization

Prime factorization is a method where a number is expressed as a product of its prime numbers. It's a fundamental concept in number theory that simplifies many problems dealing with divisors and multiples.
In our exercise, each value of \( y \) from the set \( A \) needs to be expressed in terms of its prime factors. For example, for \( y = 30 \), the prime factorization is \( 2^1 \cdot 3^1 \cdot 5^1 \). This means that 30 can be broken down into the product of three distinct prime numbers, each raised to the power of 1.
The usefulness of prime factorization here lies in its ability to systematically allow for the distribution of these distinct prime powers across our chosen variables. When we distribute these powers, we essentially count the various ways to "split" them among \( x_1, x_2, \) and \( x_3 \), leading us to the solution counts.

Binomial Coefficients

The concept of binomial coefficients helps solve combinatorial problems concerning selections and arrangements. When applied to our problem, binomial coefficients facilitate the computation of solutions for the equation \( a + b + c = n \), where \( n \) is a sum of exponents in the factorization.
The binomial coefficient \( \binom{n + 2}{2} \) gives the number of non-negative integer solutions for this equation. In our specific case in the exercise, since the power of each prime factor is 1, \( n = 1 \), the formula becomes \( \binom{1 + 2}{2} = 3 \). Therefore, for each set of prime factors of \( y \), the number of ways to arrange these factors among three variables is 3.
By calculating this for each set of prime factors separately, and due to our established relevance that each power is consistently one, the total number of solutions is the product \( 3 \times 3 \times 3 = 27 \). It's a practical example of using binomial coefficients to handle solution counting efficiently.