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 _________.

We are given the set $$S=\{1,2,3,4,5,6,7\}$$ and we must count all functions $$f:S\to S$$ that satisfy the multiplicative rule

$$f(m\cdot n)=f(m)\cdot f(n)$$

for every pair $$m,n\in S$$ for which the product $$m\cdot n$$ is still inside $$S$$. We begin by looking at the special element $$1$$.

Taking $$m=1$$ in the defining property gives the formula

$$f(1\cdot n)=f(1)\cdot f(n).$$

Because $$1\cdot n=n$$, this simplifies to

$$f(n)=f(1)\,f(n)\quad\text{for every }n\in S.$$

The only way this can hold for all $$n$$ (with the values always lying in $$S$$ and hence never being $$0$$) is when

$$f(1)=1.$$

So the value at $$1$$ is completely fixed.

Next we list all products inside $$S$$ that are still in $$S$$. Besides the trivial $$1\cdot n=n$$, only three distinct products appear:

$$2\cdot2=4,\qquad 2\cdot3=6,\qquad 3\cdot2=6.$$

No other product of two elements of $$S$$ stays within $$S$$, because, for instance, $$3\cdot3=9\notin S$$ and $$2\cdot4=8\notin S$$. Therefore the multiplicative rule imposes conditions only on the numbers $$4$$ and $$6$$. Let us introduce the convenient names

$$a=f(2),\qquad b=f(3).$$

Using the rule on each admissible product, we obtain:

1. For $$2\cdot2=4$$ we must have

$$f(4)=f(2\cdot2)=f(2)^2=a^2.$$ 2. For $$2\cdot3=6$$ (and equally for $$3\cdot2=6$$) we must have

$$f(6)=f(2\cdot3)=f(2)\,f(3)=ab.$$

Thus the values of $$f$$ at $$4$$ and $$6$$ are forced once we pick $$a=f(2)$$ and $$b=f(3)$$. In contrast, the values at $$5$$ and $$7$$ are not tied to any product lying in $$S$$, so they may be chosen freely from $$\{1,2,3,4,5,6,7\}$$ later on.

The crucial point is that every value of the function must lie in $$S$$. Hence we need

$$a^2\le 7\quad\text{and}\quad ab\le 7.$$

We now enumerate all possibilities for $$a$$ and $$b$$ satisfying these two inequalities.

• If $$a=1$$, then $$a^2=1\le7$$ automatically, and $$ab=b\le7$$ for every $$b\in\{1,2,3,4,5,6,7\}$$. So with $$a=1$$, we have $$7$$ admissible choices for $$b$$.

• If $$a=2$$, then $$a^2=4\le7$$ is still allowed, but the inequality $$ab\le7$$ becomes $$2b\le7$$, i.e. $$b\le3$$. Hence $$b$$ can be $$1,2,$$ or $$3$$, giving $$3$$ choices.

• If $$a\ge3$$, then $$a^2\ge9>7$$, which violates $$a^2\le7$$, so no further values of $$a$$ are possible.

Combining the two admissible cases, we have in total

$$7+3=10$$

legal pairs $$(a,b)=(f(2),f(3)).$$

After fixing such a pair, the values at $$4$$ and $$6$$ are already determined by $$f(4)=a^2$$ and $$f(6)=ab$$, and they automatically lie in $$S$$ by the way we chose $$a$$ and $$b$$. The final freedom rests with $$f(5)$$ and $$f(7)$$, each of which can be any of the $$7$$ members of $$S$$. Therefore the total number of distinct functions is

$$10\;\times\;7\;\times\;7 \;=\;10\;\times\;49 \;=\;490.$$

So, the answer is $$490$$.