Question

Let A = {1, 2, 3, 5, 8, 9}. Then the number of possible functions $f: A \rightarrow A$ 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 .

Answer 1

433

Answer 2

The given condition is $f(m \cdot n) = f(m) \cdot f(n)$ for $m, n, m \cdot n \in A$. The set is $A = \{1, 2, 3, 5, 8, 9\}$.

1. Consider $m=1$ or $n=1$: For any $x \in A$, $1 \cdot x = x \in A$. So, $f(x) = f(1) \cdot f(x)$. This implies either $f(x)=0$ for all $x \in A$, or $f(1)=1$.

   • Case 1: $f(1) = 0$. If $f(1)=0$, then for any $x \in A$, $f(x) = f(1) \cdot f(x) = 0 \cdot f(x) = 0$. So, $f(x)=0$ for all $x \in A$. This is one valid function.
   • Case 2: $f(1) = 1$. This condition $f(x) = f(1)f(x)$ becomes $f(x)=f(x)$, which gives no further restriction on $f(x)$ for $x \neq 1$.

2. Consider other products in A: We list all pairs $(m, n)$ such that $m, n \in A$ and $m \cdot n \in A$. The only non-trivial product is $3 \cdot 3 = 9$. Both $3$ and $9$ are in $A$. So, $f(9) = f(3) \cdot f(3) = (f(3))^2$.

3. Analyze $f(9) = (f(3))^2$: Since $f: A \rightarrow A$, $f(3)$ must be an element of $A$, and $(f(3))^2$ must also be an element of $A$. Let $y = f(3)$. Then $y \in \{1, 2, 3, 5, 8, 9\}$. We require $y^2 \in \{1, 2, 3, 5, 8, 9\}$.

   • $1^2 = 1 \in A$. So $f(3)=1$ is possible, which implies $f(9)=1$.
   • $3^2 = 9 \in A$. So $f(3)=3$ is possible, which implies $f(9)=9$. Other squares ($2^2=4, 5^2=25, 8^2=64, 9^2=81$) are not in $A$. Thus, if $f(1)=1$, $f(3)$ can only be $1$ or $3$.

4. Count the functions for $f(1)=1$:

   • Subcase 2a: $f(1)=1$ and $f(3)=1$. This implies $f(9) = (f(3))^2 = 1^2 = 1$. The values $f(1)=1$, $f(3)=1$, $f(9)=1$ are fixed. For the remaining elements $x \in \{2, 5, 8\}$, $f(x)$ can be any element of $A$. There are 6 choices for $f(2)$, 6 for $f(5)$, and 6 for $f(8)$. Number of functions = $1 \times 1 \times 6 \times 6 \times 6 = 216$.
   • Subcase 2b: $f(1)=1$ and $f(3)=3$. This implies $f(9) = (f(3))^2 = 3^2 = 9$. The values $f(1)=1$, $f(3)=3$, $f(9)=9$ are fixed. Similarly, $f(2), f(5), f(8)$ can be any element of $A$. Number of functions = $1 \times 1 \times 6 \times 6 \times 6 = 216$.

5. Total Number of Functions: Total = (Functions with $f(1)=0$) + (Functions with $f(1)=1, f(3)=1$) + (Functions with $f(1)=1, f(3)=3$) = 1 + 216 + 216 = 433.