Set A has m elements and Set B has n elements. If the total number of subsets of A is 112 more than

Q: Why did we assume $m > n$?

The problem states that the number of subsets of $A$ is $112$ *more* than the number of subsets of $B$. Since the number of subsets is $2^k$, a larger number of subsets implies a larger number of elements. Thus, $2^m > 2^n$ forces $m > n$.

Q: Is there any other way to factor $112$?

While $112$ can be written as $2 \times 56$ or $4 \times 28$, we specifically look for the form $2^n \times (\text{odd number})$ because the term $(2^{m-n} - 1)$ must be odd. This unique structure allows us to equate the parts directly.

Q: What if $m$ and $n$ were not integers?

In the context of set theory, the number of elements $m$ and $n$ must be non-negative integers. If they weren't, the concept of 'number of subsets' as $2^k$ would not be defined in the standard sense.

Q: Could $2^{m-n} - 1$ be equal to $1$?

If $2^{m-n} - 1 = 1$, then $2^{m-n} = 2$, meaning $m-n = 1$. If we substitute this back, we get $2^n(1) = 112$, which means $2^n = 112$. Since $112$ is not a power of $2$, this case is impossible.

Q: Why do we equate the odd parts?

Because $2^n$ is a power of $2$ (always even for $n \ge 1$) and $(2^{m-n} - 1)$ is always odd. By the Fundamental Theorem of Arithmetic, the prime factorization of $112$ is unique, allowing us to map the powers of $2$ and the odd factors one-to-one.

Question

Set $A$ has $m$ elements and Set $B$ has $n$ elements. If the total number of subsets of $A$ is 112 more than the total number of subsets of $B$, then the value of $m \cdot n$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm. Today, we are not just solving a problem; we are peeling back the layers of exponential growth.

Imagine you have two sets, $A$ and $B$, with $m$ and $n$ elements respectively. The number of subsets of a set is a fundamental concept in combinatorics, governed by the elegant power of two: $2^k$.

When we say set $A$ has $m$ elements, we are essentially saying there are $2^m$ ways to choose a subset from it.

### The Equation of Difference

We are told that the difference between the subsets of $A$ and $B$ is $112$. Mathematically, this gives us the beautiful, yet daunting, equation:

$$2^m - 2^n = 112$$

At first glance, this looks like a single equation with two unknowns—a situation that usually makes us nervous. But wait! We are dealing with integers, and more importantly, we are dealing with powers of $2$.

This is a hidden constraint that turns a difficult problem into a solvable puzzle.

### The Art of Factoring

To unlock this, we must use the power of factoring. Let us pull out the smaller term, $2^n$, from the expression:

$$2^n(2^{m-n} - 1) = 112$$

Now, look closely at what we have created. On the left, we have a product of an even number ($2^n$) and an odd number ($2^{m-n} - 1$).

On the right, we have the number $112$. To solve this, we must perform a prime factorization of $112$. We know that $112 = 16 	imes 7$, which is the same as $2^4 	imes 7$.

### The Elegant Comparison

This is where the magic happens. We have:

$$2^n(2^{m-n} - 1) = 2^4 	imes 7$$

Because $2^n$ is a power of $2$ and $(2^{m-n} - 1)$ is an odd number, we can equate the even and odd parts separately. By comparing the powers of $2$, we immediately see that $2^n = 2^4$, which gives us $n = 4$.

Now, for the odd part:

$$2^{m-n} - 1 = 7$$

Adding $1$ to both sides, we get $2^{m-n} = 8$. Since $8$ is $2^3$, we conclude that $m - n = 3$.

Substituting our known value of $n = 4$, we find that $m - 4 = 3$, leading us to $m = 7$.

### The Final Triumph

We have navigated the exponential landscape and arrived at our destination. We found $m = 7$ and $n = 4$.

The problem asks for the product $m \cdot n$. Thus:

$$m \cdot n = 7 	imes 4 = 28$$

Take a moment to appreciate the symmetry here. We didn't need complex calculus or heavy machinery; we relied on the fundamental properties of numbers.

When you face such problems in the JEE, remember: look for the structure, factorize with intent, and trust the logic of the integers. You have mastered this concept today, and that is a victory worth celebrating. The final answer is **28**.

Animated Solution for Mathematics - Sets and Relations: Set $A$ has $m$ elements and Set $B$ has $n$ elements. If the total number of subsets of $A$ is 112 more than the total number of subsets of $B$ , then the value of $m \cdot n$ is

Enter Numerical Value:

Visualized Solution

Visualizing the Sets

The Subset Formula

Forming the Equation

Algebraic Factoring

Prime Factorization of 112

Solving for n

Solving for m

Final Calculation

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Setup

The Equation of Difference

The Art of Factoring

The Elegant Comparison

The Final Triumph

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