Animated Solution for Mathematics - Functions: Let $S=\{1,2,3,4,5,6,7\}$. Then the number of possible functions $f:S\to S$ 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

Welcome, my dear student. Today, we are going to peel back the layers of a beautiful problem involving functional equations. These problems are not just about calculation; they are about understanding the hidden structure of numbers.

We are given a set $S=\{1,2,3,4,5,6,7\}$ and a function $f:S\to S$ that respects the multiplicative property $f(m\cdot n)=f(m)\cdot f(n)$ whenever $m,n,mn\in S$. Let us embark on this journey to count how many such functions exist.

The Foundation

Every journey begins with a single step, and in functional equations, that step is almost always finding the value of $f(1)$. Let us test the waters.

If we set $m=1$ and $n=1$, our condition becomes $f(1\cdot 1)=f(1)\cdot f(1)$, which simplifies to $f(1)=[f(1){]}^2$. This is a simple algebraic trap!

The solutions are $f(1)=0$ or $f(1)=1$. But look at our set $S=\{1,2,3,4,5,6,7\}$. The value $0$ is not invited to this party. Therefore, we must conclude that $f(1)=1$. This is our anchor.

The Chain Reaction

Now, let us look at the numbers that are linked. Consider $f(2)$.

We know that $f(4)=f(2\cdot 2)=f(2)\cdot f(2)=[f(2){]}^2$. Since the output $f(4)$ must reside within our set $S$, the square of $f(2)$ cannot exceed $7$.

Let us test the possibilities: if $f(2)=1$, then $f(4)=1^2=1$, which is in $S$. If $f(2)=2$, then $f(4)=2^2=4$, which is also in $S$. But if $f(2)=3$, then $f(4)=9$, which is outside our set.

Thus, $f(2)$ can only be $1$ or $2$. This is the fork in the road where our calculation splits into two distinct cases.

The Constraint of $f(6)$

We must also consider $f(6)$. Since $6=2\cdot 3$, we have $f(6)=f(2)\cdot f(3)$. This value must also be in $S$, meaning $f(2)\cdot f(3)\le 7$.

If $f(2)=1$, then $f(6)=1\cdot f(3)=f(3)$. Since $f(3)$ can be any value in $S$ and $f(6)$ will automatically be equal to it (and thus in $S$), $f(3)$ has $7$ choices.

The independent variables $f(5)$ and $f(7)$ also have $7$ choices each. So, for this case, we have:

If $f(2)=2$, then $f(6)=2\cdot f(3)$. For $f(6)$ to be in $S$, we need $2\cdot f(3)\le 7$, which means $f(3)$ can only be $1,2,$ or $3$.

That gives us $3$ choices for $f(3)$. Again, $f(5)$ and $f(7)$ remain free with $7$ choices each. So, for this case, we have:

The Synthesis

We have navigated the dependencies and identified the free agents. The total number of functions is simply the sum of our two cases:

It is elegant, isn't it? By breaking the problem down into logical dependencies, we turned a daunting functional equation into a simple counting exercise.

Keep practicing this systematic approach, and you will find that even the most complex JEE problems start to reveal their secrets. The final answer is 490.