Let $S=\{1,2,3,4,5,6,7\}$. Then the number of possible functions $\mathrm{f}: \mathrm{S} \rightarrow

Question

Let $S=\{1,2,3,4,5,6,7\}$. Then the number of possible functions $\mathrm{f}: \mathrm{S} \rightarrow \mathrm{S}$ such that $\mathrm{f}(\mathrm{m} \cdot \mathrm{n})=\mathrm{f}(\mathrm{m}) \cdot \mathrm{f}(\mathrm{n})$ for every $\mathrm{m}, \mathrm{n} \in \mathrm{S}$ and $\mathrm{m} \cdot \mathrm{n} \in \mathrm{S}$ is equal to $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

Given,
$$
\mathrm{F}(\mathrm{mn})=\mathrm{f}(\mathrm{~m}) \cdot \mathrm{f}(\mathrm{n})
$$
Put $\mathrm{m}=1$, then $\mathrm{f}(1 \cdot \mathrm{n})=\mathrm{f}(1) \cdot \mathrm{f}(\mathrm{n}) \Rightarrow \mathrm{f}(1)=1 \quad\{f(n) 
eq 0\}$
Put $\mathbf{m}=\mathbf{n}=2$
$$
f(4)=f(2) \cdot f(2)
$$
If $f(2)=1$, then $f(4)=1$ and if $f(2)=2$, then $f(4)=4$
Put $\mathrm{m}=2, \mathrm{n}=3$
$$
f(6)=f(2) \cdot f(3)
$$
If $f(2)=1$, then $f(3)$ can be any value between 1 to 7
If $f(2)=2$, then $f(3)$ can be 1 or 2 or 3
$f(5), f(7)$ can take any value from $\{1,2, \ldots, 7\}$
So, total number of function $=(1 	imes 1 	imes 7 	imes 1 	imes 7 	imes 1 	imes 7)+(1 	imes 1 	imes 3 	imes 1 	imes 7 	imes 1 	imes 7)$
$$
=490
$$