Let A = 1, 2, 3, 5, 8, 9 . Then the number of possible functions f:...

Given,

$f\left(m \cdot n\right) = f\left(m\right) \cdot f\left(n\right)$

Taking $m = 1$

$⇒ f\left(1 \cdot n\right) = f\left(1\right) \cdot f\left(n\right)$

$⇒ f\left(1\right) = 1$

Now taking $m = 3 \text{ and } n = 3$ we get,

$f ( 9 ) = f ( 3 ) \times f ( 3 )$,

So, \left(f\left(3\right), f\left(9\right)\right)$have two possibility \left(1, 1\right) \text{ and } \left(3, 9\right)$

Now taking $m = 2 \text{ and } n = 1$ we get,

$f\left(2 \cdot 1\right) = f\left(2\right) \cdot f\left(1\right)$

Now here $f\left(1\right) = 1$, so $f\left(2\right)$ can take all $6$ numbers,

Similarly, for $f\left(5\right) \text{ and } f\left(8\right)$ there will be $6$ ways each,

So, total possible function will be, $= 1 \times 6 \times 2 \times 6 \times 6 \times 1 = 432$