To solve the problem, let's break it down step by step.

Given:

1. The number of subsets of a finite set with $k$ elements is $2^k$.
2. For two sets with $m$ and $n$ elements, the total number of subsets of the first set is $2^m$, and the total number of subsets of the second set is $2^n$.
3. It is given that: $2^m = 2^n + 56$

Step 1: Express the relationship

From the equation $2^m = 2^n + 56$, rearrange: $2^m - 2^n = 56$ Factorize the left-hand side using the property $2^m - 2^n = 2^n(2^{m-n} - 1)$: $2^n(2^{m-n} - 1) = 56$

Step 2: Solve for $n$ and $m-n$

1. Since $2^n$ is a power of 2, and $56 = 2^3 \cdot 7$, $2^n$ must divide $56$. Thus, possible values of $2^n$ are $1, 2, 4, 8, 16,$ etc., but only those $\leq 56$: $1, 2, 4, 8$.

2. For each value of $2^n$, solve $2^n(2^{m-n} - 1) = 56$ to check for integer solutions.

Case 1: $2^n = 8$ ($n = 3$)

$8(2^{m-3} - 1) = 56$ Divide by 8: $2^{m-3} - 1 = 7$ $2^{m-3} = 8$ $m-3 = 3 \implies m = 6$

Case 2: $2^n = 4$ ($n = 2$)

$4(2^{m-2} - 1) = 56$ Divide by 4: $2^{m-2} - 1 = 14$ $2^{m-2} = 15$ But $15$ is not a power of 2, so this case fails.

Case 3: $2^n = 2$ ($n = 1$)

$2(2^{m-1} - 1) = 56$ Divide by 2: $2^{m-1} - 1 = 28$ $2^{m-1} = 29$ But $29$ is not a power of 2, so this case fails.

Step 3: Verify solution

The only solution is: $m = 6, \, n = 3$

Final Answer:

The values of $m$ and $n$ are: $m = 6, \, n = 3$

Would you like to see a detailed explanation of the verification process or alternate approaches?

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Related Questions:

1. How many total subsets does a set with $n$ elements have?
2. What is the relationship between $2^m$ and $2^n$ in this problem?
3. If the difference between the subsets of two sets were 120, what would be the new values of $m$ and $n$?
4. How is the property $2^a - 2^b = 2^b(2^{a-b} - 1)$ useful in such problems?
5. Can this problem be generalized to sets with infinite elements?

Tip:

Always express powers of two explicitly and use factorization techniques for solving equations involving $2^k$.