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Step-by-Step Solution
Step 1: Understand the Functional Condition
The problem states that for every pair of elements m, n \in A such that their product m \cdot n is also in A , the function f satisfies:
f(m \cdot n) = f(m) \cdot f(n).
We need to determine how many such functions from A to A exist, where A = \{1, 2, 3, 5, 8, 9\} .
Step 2: Determine f(1)
Substitute m = 1 in the given condition. For any n \in A , we have:
f(1 \cdot n) = f(1) \cdot f(n).
Since 1 \cdot n = n , this becomes:
f(n) = f(1) \cdot f(n).
This equation can only hold for all n if f(1) = 1.
Step 3: Consider f(3) and f(9)
Because 9 = 3 \cdot 3 and 9 \in A , the functional equation gives us:
f(9) = f(3 \cdot 3) = f(3) \cdot f(3) = \bigl(f(3)\bigr)^2.
This implies that f(9) must be the square of f(3). However, f(3) and f(9) can each be any element of A. The only way f(9) can match the square of f(3) (and still lie in A ) is if:
f(3) = 1 and f(9) = 1, or
f(3) = 3 and f(9) = 9.
Therefore, there are 2 possible assignments for the pair (f(3), f(9)).
Step 4: Assign Values to f(2) , f(5) , and f(8)
For the remaining elements 2, 5, and 8, there is no further restriction from the given condition aside from the fact that the function values must lie in A. Each of f(2), f(5), f(8) can independently take any of the 6 values in A = \{1, 2, 3, 5, 8, 9\}.
Hence, there are 6 choices for f(2) , 6 choices for f(5) , and 6 choices for f(8) . In total, for these three elements:
6 \times 6 \times 6 = 6^3 = 216 possible ways.
Step 5: Combine All Possibilities
We have:
f(1) is fixed as 1 (1 way).
(f(3), f(9)) can be chosen in 2 ways.
f(2), f(5), f(8) can be chosen in 6^3 = 216 ways.
Therefore, the total number of such functions is:
2 \times 216 = 432.
Final Answer
\boxed{432}
