Let A = {1, 2, 3, 5, 8, 9}. Then the number o | vedclass

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(B) Given $A = \{1, 2, 3, 5, 8, 9\}$. The condition is $f(m \cdot n) = f(m) \cdot f(n)$ whenever $m, n, m \cdot n \in A$.$1$. For $m=1, n=1$: $f(1) =

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$432$

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(B) Given $A = \{1, 2, 3, 5, 8, 9\}$. The condition is $f(m \cdot n) = f(m) \cdot f(n)$ whenever $m, n, m \cdot n \in A$.$1$. For $m=1, n=1$: $f(1) = f(1) \cdot f(1) \implies f(1) = 1$ (since $f(1) \in A$ and $f(1) 
eq 0$).$2$. For $m=3, n=3$: $f(9) = f(3) \cdot f(3) = (f(3))^2$. Since $f(9) \in A$,$(f(3))^2$ must be in $A$. Possible values for $f(3)$ are $1$ (as $1^2=1 \in A$) or $3$ (as $3^2=9 \in A$).$3$. For $m=2, n=4$ (not in $A$),we look at available products: $2 \cdot 4$ is not in $A$. There are no constraints on $f(2), f(5), f(8)$ other than they must map to elements in $A$.$4$. The values $f(1)$ and $f(9)$ are determined by $f(3)$.- If $f(3) = 1$,then $f(9) = 1^2 = 1$. ($1$ choice for $f(3)$)- If $f(3) = 3$,then $f(9) = 3^2 = 9$. ($1$ choice for $f(3)$)$5$. The elements $f(2), f(5), f(8)$ can be any of the $6$ elements in $A$ because there are no products $m \cdot n$ in $A$ involving these values that impose further restrictions.Total functions = (Choices for $f(3)$) $	imes$ (Choices for $f(2)$) $	imes$ (Choices for $f(5)$) $	imes$ (Choices for $f(8)$)$= 2 	imes 6 	imes 6 	imes 6 = 432$.

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 $...............$.

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