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Step-by-Step Solution
Step 1: Understanding the Functional Equation
We have a function f: S \to S where S = \{1,2,3,4,5,6,7\} such that for all m,n \in S with m \cdot n \in S ,
f(m \cdot n) = f(m) \cdot f(n) .
We need to find the number of such possible functions f .
Step 2: Determine f(1)
Let m = 1 . Then the functional equation becomes:
f(1 \cdot n) = f(1) \cdot f(n) \implies f(n) = f(1) \cdot f(n) .
For this to hold for every n \in S , we must have f(1) = 1 .
Step 3: Possible Values for f(2) and f(4)
Set m = n = 2 . Then:
f(4) = f(2 \cdot 2) = f(2) \cdot f(2) = [f(2)]^2.
Since f(4) must be in S = \{1,2,3,4,5,6,7\} , the square of f(2) must also lie in S . The possible integer squares in S are 1 and 4 . Thus:
If f(2) = 1 , then f(4) = 1.
If f(2) = 2 , then f(4) = 4.
Step 4: Determining f(3) and f(6)
Now consider m = 2, n = 3 (so their product is 6 ):
f(6) = f(2) \cdot f(3).
Depending on f(2) , the choices for f(3) and hence f(6) vary:
If f(2) = 1 , then f(6) = 1 \cdot f(3) = f(3) . Here f(3) can be any value from 1 to 7 .
If f(2) = 2 , then f(6) = 2 \cdot f(3) . For f(6) to stay in S , f(3) can be 1, 2, or 3 (so that 2 \cdot f(3) remains in S ).
Step 5: Possible Assignments to f(5) and f(7)
Notice that 5 and 7 do not appear as products of smaller numbers in S that stay within S (e.g., 2.5=10 not in S ; 3.5=15 not in S ; and so on), so f(5) and f(7) can each take any of the 7 values in S independently.
Step 6: Counting All Valid Functions
We count separately for the two cases of f(2) :
Case 1: f(2) = 1
• f(4) = 1 .
• f(3) can be any of 7 values in S .
• f(6) = f(2)\cdot f(3) = 1 \cdot f(3) = f(3) (no further restriction).
• f(5) can be any of 7 values.
• f(7) can be any of 7 values.
Total possibilities in this case = 1 \times 1 \times 7 \times 7 \times 7 = 343.
Case 2: f(2) = 2
• f(4) = 4 .
• f(3) can be 1, 2, or 3 .
• f(6) = 2 \cdot f(3) must remain in S , which is already ensured by choosing f(3) as 1, 2, or 3 .
• f(5) can be any of 7 values.
• f(7) can be any of 7 values.
Total possibilities in this case = 1 \times 1 \times 3 \times 7 \times 7 = 147.
Step 7: Final Summation
Add the counts from both cases:
Total number of possible functions = 343 + 147 = 490.
Answer
490
