Problem 82 - A Complete Resource Book in Mathematics for JEE Main 2017

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 correct answer is (B) 27.

Step 1: Understanding the Problem

We are given a set \( A = \{1, 2, 3, 5, 6, 10, 15, 30\} \) and need to find the number of positive integral solutions for \( x_1 x_2 x_3 = y \) for each \( y \in A \). These solutions can be any combination of integers that multiply to \( y \).

Step 2: Total Ways to Distribute Factors

For any positive integer \( y \) and the equation \( x_1 x_2 x_3 = y \), each \( y \) can be expressed as the product of its prime factors raised to their respective powers. If \( y = p_1^{a_1} p_2^{a_2} \ldots p_n^{a_n} \), then distributing these factors among \( x_1, x_2, x_3 \) results in the formula: \((a_1 + 2)\)\((a_2 + 2)\)\ldots\((a_n + 2)\) total solutions.

Step 3: Solution Calculation for Each \( y \)

We list each \( y \) in \( A \) and apply the formula:1. \( 1 = 1^1 \): \((1+2) = 3\) solutions. 2. \( 2 = 2^1 \): \((1+2) = 3 \) solutions.3. \( 3 = 3^1 \): \((1+2) = 3 \) solutions.4. \( 5 = 5^1 \): \((1+2) = 3 \) solutions.5. \( 6 = 2^1 \, 3^1 \): \((1+2)(1+2) = 9 \) solutions.6. \( 10 = 2^1 \, 5^1 \): \((1+2)(1+2) = 9 \) solutions.7. \( 15 = 3^1 \, 5^1 \): \((1+2)(1+2) = 9 \) solutions.8. \( 30 = 2^1 \, 3^1 \, 5^1 \): \((1+2)(1+2)(1+2) = 27 \) solutions.

Step 4: Identifying the Maximum Solutions

Upon calculating the number of solutions for each \( y \), we observe that 27 is the maximum number of solutions, which occurs when \( y = 30 \). No other \( y \) leads to a greater number than 27.

Key Concepts

Integers

Integers are a foundation of mathematics and consist of zero, positive, and negative whole numbers. An integer does not have fractions or decimals; it is a complete number.
In everyday scenarios, integers help in counting objects, managing accounts, and even measuring temperature changes as they accommodate both negative and positive numbers. In our exercise, the focus is on positive integers, represented by numbers like 1, 2, 3, and so on. When working with integers in mathematical problems, their properties, such as divisibility and the fact that they are whole numbers, become crucial in forming solutions.
Remember, in equations like the one presented, where the products of integers need to equal a specific value, understanding how integers operate is key to cracking the problem effectively.

Prime Factorization

Prime factorization is the process of breaking down a number into its basic building blocks, which are prime numbers. A prime number is a number greater than 1 that has no positive divisors other than 1 and itself.
For example, 6 can be broken down into the prime factors 2 and 3. This process reveals the smallest prime numbers that multiply together to form the original number.
In the given problem, each element in the set is expressed in terms of its prime factors. This simplification helps to systematically assign the factors to different variables in the problem equation. Calculating the number of positive integral solutions involves an understanding of this decomposition since the multiplication of these factors in different combinations leads to determining the possible configurations of the equation solution.

Positive Integral Solutions

When dealing with equations like \(x_1 x_2 x_3 = y\), the goal is to find sets of integers \((x_1, x_2, x_3)\) that satisfy the equation where each variable is a positive integer (i.e., greater than zero).
From a combinatorial perspective, each positive integer solution represents a different permutation or combination of the factors that result in the original number. For instance, if \(y\) equals 6 with prime factorization of \(2^1 \, 3^1\), the combinations \((x_1, x_2, x_3)\) that could equal this multiplication involve distributing these powers among the three variables.
The exercise simplifies finding all possible distributions by employing formulas derived from combinatorics, which is a branch of mathematics focused on counting, arrangement, and combination of elements within a set. By doing so, we can systematically determine all possible configurations without manually listing each triple \((x_1, x_2, x_3)\). This makes the problem more efficient to solve.