QJEEM21Mab4e9: JEE Main 2021 Mathematics - Class 11 JEEM21Mab4e9 Solution: Let S = {1, 2, 3, 4, 5, 6, 7}. Then the number of possible…

F(mn) = f(m) . f(n)

Put m = 1 f(n) = f(1) . f(n) $\Rightarrow$ f(1) = 1

Put m = n = 2

$$f(4)=f(2).f(2)\{\begin{array}{c}f(2)=1\Rightarrow f(4)=1\hfill \\ or\hfill \\ f(2)=2\Rightarrow f(4)=4\hfill \end{array}$$

Put m = 2, n = 3

$$f(6)=f(2).f(3)\{\begin{array}{c}whenf(2)=1\hfill \\ f(3)=1to7\hfill \\ \\ f(2)=2\hfill \\ f(3)=1or2or3\hfill \end{array}$$

f(5), 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