Solve this following

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 $m, n \in S$ and $m \cdot n \in S$ is equal to

Solution:

$\mathrm{F}(\mathrm{mn})=\mathrm{f}(\mathrm{m}) \cdot \mathrm{f}(\mathrm{n})$

Put $\mathrm{m}=1 \mathbf{f}(\mathbf{n})=\mathbf{f}(\mathbf{1}) \cdot \mathrm{f}(\mathrm{n}) \Rightarrow \mathrm{f}(1)=1$

Put $\mathrm{m}=\mathrm{n}=2$

$f(4)=f(2) \cdot f(2)\left\{\begin{array}{l}f(2)=1 \Rightarrow f(4)=1 \\ \text { or } \\ f(2)=2 \Rightarrow f(4)=4\end{array}\right.$

Put $m=2, n=3$

$f(6)=f(2) . f(3)\left\{\begin{array}{l}\text { when } f(2)=1 \\ f(3)=1 \text { to } 7 \\ f(2)=2 \\ f(3)=1 \text { or } 2 \text { or3 }\end{array}\right.$

$\mathrm{f}(5), \mathrm{f}(7)$ can take any value

Total $=(1 \times 1 \times 7 \times 1 \times 7 \times 1 \times 7)$

$+(1 \times 1 \times 3 \times 1 \times 7 \times 1 \times 7)$

$=490$

 

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