The total number of subsets of a finite set A has 56 more elements than the total number of subsets of another finite set B. What is the number of elements in the set A?

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the problem, we need to find the number of elements in set A given that the total number of subsets of set A has 56 more elements than the total number of subsets of set B.

### Step-by-Step Solution:

1. **Define the Variables**:
Let the number of elements in set A be \( M \) and the number of elements in set B be \( N \).

2. **Calculate the Number of Subsets**:
The total number of subsets of a set with \( M \) elements is given by \( 2^M \).
Similarly, the total number of subsets of a set with \( N \) elements is \( 2^N \).

3. **Set Up the Equation**:
According to the problem, the number of subsets of set A is 56 more than the number of subsets of set B. This can be expressed as:
\[ 2^M = 2^N + 56 \]

4. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ 2^M - 2^N = 56 \]

5. **Factor the Left Side**:
We can factor the left side:
\[ 2^N(2^{M-N} - 1) = 56 \]

6. **Finding Possible Values**:
We need to find integer values for \( N \) and \( M \) such that \( 2^N(2^{M-N} - 1) = 56 \). We can start by finding the factors of 56:
The factors of 56 are \( 1, 2, 4, 7, 8, 14, 28, 56 \).

7. **Testing Values for \( N \)**:
- If \( N = 3 \):
\[ 2^3(2^{M-3} - 1) = 56 \implies 8(2^{M-3} - 1) = 56 \implies 2^{M-3} - 1 = 7 \implies 2^{M-3} = 8 \implies M - 3 = 3 \implies M = 6 \]
- If \( N = 4 \):
\[ 2^4(2^{M-4} - 1) = 56 \implies 16(2^{M-4} - 1) = 56 \implies 2^{M-4} - 1 = 3 \implies 2^{M-4} = 4 \implies M - 4 = 2 \implies M = 6 \]
- If \( N = 5 \):
\[ 2^5(2^{M-5} - 1) = 56 \implies 32(2^{M-5} - 1) = 56 \implies 2^{M-5} - 1 = 1 \implies 2^{M-5} = 2 \implies M - 5 = 1 \implies M = 6 \]

8. **Conclusion**:
The only valid solution is \( M = 6 \) and \( N = 3 \). Thus, the number of elements in set A is:
\[ \boxed{6} \]