Let S = 1, 2, 3, 4, 5, 6, 7. Then the number of possible functions $f: S \rightarrow S$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in S$ and $m \cdot n \in S$ is equal to _________.
Show Hint
For functions with a multiplicative property, first identify the prime numbers in the domain. The function's values on these primes often determine its values on the composite numbers. Then, systematically count the valid choices for the prime images, ensuring the dependent images remain in the codomain.
Step 1: Value of \(f(1)\)
Put \(m=1\). Since \(1\cdot n=n\in S\),
\[
f(n)=f(1\cdot n)=f(1)\,f(n)
\]
As \(f(n)\neq 0\), we must have
\[
f(1)=1
\]
Step 2: Identify product relations in \(S\)
All non-trivial products in \(S\) that remain in \(S\) are:
\[
2\cdot2=4,\qquad 2\cdot3=6
\]
Thus, the functional conditions are:
\[
f(4)=f(2)^2,\qquad f(6)=f(2)f(3)
\]
Step 3: Independent and dependent values
The prime elements in \(S\) are:
\[
2,\;3,\;5,\;7
\]
Values of \(f(5)\) and \(f(7)\) are completely free (no constraints).
Values of \(f(4)\) and \(f(6)\) depend on \(f(2)\) and \(f(3)\).
Step 4: Determine possible values of \(f(2)\)
Since \(f(4)=f(2)^2\in S\):
\[
f(2)=1 \Rightarrow f(4)=1 \in S \quad (\text{valid})
\]
\[
f(2)=2 \Rightarrow f(4)=4 \in S \quad (\text{valid})
\]
\[
f(2)\ge 3 \Rightarrow f(2)^2\ge 9 \notin S \quad (\text{invalid})
\]
Hence,
\[
f(2)\in\{1,2\}
\]
Step 5: Count possibilitiesCase I: \(f(2)=1\)
Then
\[
f(6)=f(2)f(3)=f(3)\in S
\]
So, \(f(3)\) has 7 choices.
Number of choices in this case:
\[
7
\]
Case II: \(f(2)=2\)
Then
\[
f(6)=2f(3)\in S
\]
Valid values of \(f(3)\):
\[
f(3)=1\Rightarrow f(6)=2
\]
\[
f(3)=2\Rightarrow f(6)=4
\]
\[
f(3)=3\Rightarrow f(6)=6
\]
So, \(f(3)\) has 3 choices.
Number of choices in this case:
\[
3
\]
Step 6: Free choices
\[
f(5)\in S \Rightarrow 7 \text{ choices}
\]
\[
f(7)\in S \Rightarrow 7 \text{ choices}
\]
Total number of functions
\[
(7+3)\times7\times7=490
\]
Answer: \(\boxed{490}\)
