Question

Let \( A = \{1, 2, 3, \ldots, 7\} \) and let \( P(A) \) denote the power set of \( A \). If the number of functions \( f : A \rightarrow P(A) \) such that \( a \in f(a), \, \forall a \in A \) is \( m^n \), \( m \) and \( n \in \mathbb{N} \) and \( m \) is least, then \( m + n \) is equal to \(\_\_\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 44

The problem asks for the value of \(m + n\), where \(m^n\) is the total number of functions \(f: A \to P(A)\) satisfying the condition \(a \in f(a)\) for all \(a \in A\), with \(A = \{1, 2, 3, \ldots, 7\}\) and \(m\) being the least possible natural number.

Concept Used:

The solution involves combinatorial counting principles. A function \(f: A \to B\) is defined by assigning to each element \(a \in A\) exactly one element from \(B\). The total number of such functions is \(|B|^{|A|}\).

In this problem, the domain is \(A\) and the codomain is the power set of A, \(P(A)\). A special condition is imposed on the function, which restricts the possible outputs for each input. To find the total number of such functions, we need to determine the number of valid choices for the output \(f(a)\) for each input \(a \in A\) and then multiply these numbers together, as the choice for each input is independent.

Step-by-Step Solution:

Step 1: Analyze the domain and codomain of the function.

The domain is the set \(A = \{1, 2, 3, 4, 5, 6, 7\}\). The size of the domain is \(|A| = 7\).

The codomain is the power set of A, \(P(A)\), which is the set of all subsets of A. The size of the power set is \(|P(A)| = 2^{|A|} = 2^7\).

A function \(f: A \to P(A)\) maps each element \(a \in A\) to a subset of A, denoted by \(f(a)\).

Step 2: Apply the given condition to determine the number of choices for each function value.

The condition is that for any element \(a \in A\), we must have \(a \in f(a)\). This means that for each \(a\), the subset \(f(a)\) that the function assigns must contain the element \(a\) itself.

Let's find the number of possible subsets for an arbitrary element, say \(a_k \in A\). The subset \(f(a_k)\) must be a subset of \(A\) and must contain \(a_k\).

A subset of \(A\) is formed by deciding for each of the 7 elements of \(A\) whether to include it or not. For the subset \(f(a_k)\):

• The inclusion of the element \(a_k\) is fixed (it must be in the set). There is only 1 choice for this element.
• For the other \(|A| - 1 = 7 - 1 = 6\) elements in \(A\), each can either be included in \(f(a_k)\) or not. This gives 2 choices for each of these 6 elements.

Therefore, the total number of valid subsets that can be assigned to \(a_k\) is \(1 \times 2^6 = 64\).

Step 3: Calculate the total number of such functions.

The choice of the image \(f(a)\) for each \(a \in A\) is independent. The total number of functions is the product of the number of choices for each element in the domain.

Since \(|A| = 7\), and for each of the 7 elements there are \(2^6\) possible images, the total number of functions is:

\[ \text{Total functions} = \underbrace{(2^6) \times (2^6) \times \cdots \times (2^6)}_{7 \text{ times}} = (2^6)^7 \] \[ \text{Total functions} = 2^{6 \times 7} = 2^{42} \]

Step 4: Express the total number in the form \(m^n\) with the least \(m\).

We are given that the total number of functions is \(m^n\), where \(m, n \in \mathbb{N}\) and \(m\) is the least possible value. We have:

\[ m^n = 2^{42} \]

For \(m\) to be the least possible natural number (and \(m > 1\)), we must choose the smallest possible base, which is the prime base of the number. In this case, the base is 2.

So, we let \(m = 2\). Then, we have:

\[ 2^n = 2^{42} \implies n = 42 \]

The values are \(m = 2\) and \(n = 42\).

Final Computation & Result:

Step 5: Calculate the value of \(m + n\).

With \(m = 2\) and \(n = 42\), the sum is:

\[ m + n = 2 + 42 = 44 \]

The value of \(m + n\) is 44.

Answer 2

Each element \( a \) in \( A \) must be included in its corresponding subset in \( P(A) \), so we only consider subsets of \( A \) that contain \( a \). For each element, there are \( 2^6 \) possible subsets of \( A \) that include \( a \) (since we can select or omit any of the remaining 6 elements).

Thus, for each \( a \in A \), there are \( 2^6 \) choices, and since there are 7 elements in \( A \):

Total number of functions = \( (2^6)^7 = 2^{42} \)

Since we need \( m^n = 2^{42} \) with \( m \) and \( n \) as small as possible:

\( m = 2 \), \( n = 42 \)

Therefore, \( m + n = 2 + 42 = 44 \).