Let A = {1, 2, 3, 5, 8, 9}. Then the number of possible functions f: A → A such that f(m · n) = f(m)

Q: Wait, why do we reject the negative root when solving $f(9) = (f(3))^2$?

Great observation! We don't necessarily reject a negative root; rather, the codomain $A$ is defined as $\{1, 2, 3, 5, 8, 9\}$. Since all elements in $A$ are positive, any value $f(n)$ must be positive. Thus, negative roots are impossible by the very definition of our set $A$.

Q: Why can't we just say there are $6^6$ total functions?

That would be true if there were no constraints! The condition $f(m \cdot n) = f(m) \cdot f(n)$ acts like a filter. It forces certain elements to 'behave' in relation to others. We only use $6^6$ when every input is independent. Here, $f(9)$ is shackled to $f(3)$, reducing our total possibilities.

Q: How do I know if I missed a pair $(m, n)$?

Systematic checking is your best friend. Create a small multiplication table for the elements of $A$. If $m \cdot n$ is not in $A$, that pair doesn't impose a constraint. You will quickly see that $3 \cdot 3 = 9$ is the only non-trivial interaction.

Q: Does $f(1)$ have to be $1$?

Yes. By substituting $m=1$ into $f(m \cdot n) = f(m) \cdot f(n)$, we get $f(n) = f(1) \cdot f(n)$. Since $f(n)$ is never zero (as $0 \notin A$), we can divide both sides by $f(n)$ to get $f(1) = 1$. It is a rigid requirement of the functional equation.

Q: What if $f(3) = 5$? Why is that invalid?

If $f(3) = 5$, then the equation $f(9) = (f(3))^2$ forces $f(9) = 5^2 = 25$. But $25$ is not in our set $A$. Since the function must map to $A$, $f(9)$ must be an element of $A$. Therefore, $f(3)=5$ is impossible.

Question

Let $A = \{1, 2, 3, 5, 8, 9\}$. Then the number of possible functions $f: A 	o A$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given a set $A = \{1, 2, 3, 5, 8, 9\}$ and a functional equation $f(m \cdot n) = f(m) \cdot f(n)$, which holds for all $m, n$ such that $m \cdot n \in A$. This is a multiplicative functional equation defined over a restricted domain.

### The Anchor of Identity

Every journey begins with a single step. In functional equations, that step is testing the identity element. If we set $m = 1$, the equation transforms into:

$$f(1 \cdot n) = f(1) \cdot f(n)$$

This simplifies to $f(n) = f(1) \cdot f(n)$. Because our codomain is $A = \{1, 2, 3, 5, 8, 9\}$, we know that $f(n)$ can never be zero.

We can safely divide both sides by $f(n)$, revealing that $f(1) = 1$. Thus, there is only **$1$ choice** for $f(1)$.

### Hunting for Interactions

The rule only constrains us when $m \cdot n \in A$. Let us examine the elements of $A$ to see which pairs produce a product that remains within the set.

Most elements are "loners" that do not interact to produce a result within $A$. However, consider the element $3$. If we take $m=3$ and $n=3$, then $m \cdot n = 9$.

Since $9 \in A$, this pair creates a bridge:

$$f(9) = f(3) \cdot f(3) = (f(3))^2$$

This is the only non-trivial constraint in the entire problem. The value of $f(9)$ is entirely dependent on the choice we make for $f(3)$.

### The Logic of Possibility

We must satisfy the constraint $f(9) = (f(3))^2$, where both $f(3)$ and $f(9)$ are members of $A$. Let us test the possible values for $f(3)$:

*   If $f(3) = 1$, then $f(9) = 1^2 = 1$. Since $1 \in A$, this is **valid**.
*   If $f(3) = 3$, then $f(9) = 3^2 = 9$. Since $9 \in A$, this is **valid**.
*   If $f(3) = 2$, then $f(9) = 2^2 = 4$. Since $4 
otin A$, this is **invalid**.
*   If $f(3) = 5$, then $f(9) = 25 
otin A$. This is **invalid**.
*   If $f(3) = 8$, then $f(9) = 64 
otin A$. This is **invalid**.
*   If $f(3) = 9$, then $f(9) = 81 
otin A$. This is **invalid**.

We have exactly **$2$ valid scenarios** for the pair $(f(3), f(9))$.

### The Freedom of the Unconstrained

We have accounted for $1, 3,$ and $9$. The remaining elements, $2, 5,$ and $8$, have no constraints imposed by the functional equation.

These elements are "free agents." They can map to any of the $6$ elements in $A$ without violating the rule.

For $f(2)$, we have $6$ choices. For $f(5)$, we have $6$ choices. For $f(8)$, we have $6$ choices.

### Final Calculation

To find the total number of functions, we apply the Fundamental Counting Principle by multiplying the independent choices together:

$$	ext{Total} = (	ext{Choices for } f(1)) 	imes (	ext{Choices for } f(3), f(9)) 	imes (	ext{Choices for } f(2)) 	imes (	ext{Choices for } f(5)) 	imes (	ext{Choices for } f(8))$$

Substituting our determined values:

$$	ext{Total} = 1 	imes 2 	imes 6 	imes 6 	imes 6$$

$$	ext{Total} = 2 	imes 216 = 432$$

The total number of such functions is **$432$**.

Animated Solution for Mathematics - Functions: Let $A = {1, 2, 3, 5, 8, 9}$ . Then the number of possible functions $f : A \to A$ such that $f (m \cdot n) = f (m) \cdot f (n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Domain and Condition

The Role of $1$ in the Equation

Determining $f (1)$

Finding Valid Pairs $(m, n)$

The Special Case of $3$ and $9$

Evaluating Choices for $f (3)$ - Case 1

Evaluating Choices for $f (3)$ - Case 2

Identifying Unconstrained Elements

Counting Mappings for Free Elements

Calculating Total Number of Functions

The Sigma Insight: Classification of Functions

Analyzing the Setup

The Anchor of Identity

Hunting for Interactions

The Logic of Possibility

The Freedom of the Unconstrained

Final Calculation

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Visualized Solution

Understanding the Domain and Condition

The Role of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1 in the Equation

Determining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(1)

Finding Valid Pairs .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (m,n)

The Special Case of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 9

Evaluating Choices for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3) - Case 1

Evaluating Choices for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(3) - Case 2