1. Understand the condition

We need functions $f:S\to S$, where $S=\{1,2,3,4,5,6,7\},$ such that $f(mn)=f(m)f(n)$ for every $m,n\in S$ with $mn\in S$.

So the multiplicative property is required only when the product still lies in $S$.

2. List all products inside $S$

Since $S=\{1,2,3,4,5,6,7\}$, we only care about pairs $(m,n)$ with $mn\le 7$.

The relevant ones are:

• $1\cdot k=k$ for all $k\in S$
• $2\cdot 2=4$
• $2\cdot 3=6$
• $3\cdot 2=6$

No other nontrivial products remain in $S$ because:

• $2\cdot 4=8>7$
• $3\cdot 3=9>7$
• and larger products also exceed $7$.

Thus the only constraints are:

$f(1\cdot k)=f(1)f(k)=f(k) \quad \forall k\in S,$ $f(4)=f(2)^2,$ $f(6)=f(2)f(3).$

3. Determine $f(1)$

From $f(k)=f(1)f(k) \quad \forall k\in S,$ we get $(f(1)-1)f(k)=0 \quad \forall k\in S.$

Since $f(k)\in S=\{1,2,3,4,5,6,7\}$, none of the values can be $0$. Hence necessarily $f(1)=1.$

4. Use the constraint on $f(4)$

We need $f(4)=f(2)^2.$ Since $f(4)\in S$, we must have $f(2)^2\in S$.

Now check possible $f(2)\in S$:

• if $f(2)=1$, then $f(4)=1$
• if $f(2)=2$, then $f(4)=4$
• if $f(2)=3$, then $f(4)=9\notin S$
• larger values also give squares $>7$

So only $f(2)=1 \text{ or } 2.$

Thus there are 2 choices for $f(2)$, and then $f(4)$ is forced.

5. Use the constraint on $f(6)$

We need $f(6)=f(2)f(3).$ Also $f(6)\in S$, so $f(2)f(3)\le 7.$

We split into cases.

Case 1: $f(2)=1$

Then $f(6)=f(3).$ Now $f(3)$ can be any element of $S$, since $1\cdot f(3)=f(3)\in S$ always.

So in this case, 7 choices for $f(3)$.

Case 2: $f(2)=2$

Then $f(6)=2f(3).$ We need $2f(3)\in S$, so $2f(3)\le 7 \implies f(3)\le 3.$ Hence $f(3)\in \{1,2,3\}.$

So in this case, 3 choices for $f(3)$.

Therefore total valid choices for $(f(2),f(3))$ are $7+3=10.$

And then $f(4)$ and $f(6)$ are determined automatically.

6. Choose remaining values freely

What about $f(5)$ and $f(7)$?

There are no additional constraints involving them, because:

• $5\cdot 2,5\cdot 3,\dots >7$
• $7\cdot 2,7\cdot 3,\dots >7$
• only $1\cdot 5=5$ and $1\cdot 7=7$, which are already handled by $f(1)=1$.

So:

• $f(5)$ has 7 choices
• $f(7)$ has 7 choices

Also $f(1)=1$ is fixed, and $f(4),f(6)$ are determined by $f(2),f(3)$.

7. Total number of functions

Thus the total number is $10\times 7\times 7 = 490.$

8. Compare with stored answer

Derived answer = $490$.

Stored correct answer = $490$.

They match.