The number of functions f :{1,2,3,4} rightarr | vedclass

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Question

$f:\{1,2,3,4\} \rightarrow\{ a \in Z :| a | \leq 8\}$

$f( n )+\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$

$f( n +1)$ must be divisible by

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$f:\{1,2,3,4\} \rightarrow\{ a \in Z :| a | \leq 8\}$

$f( n )+\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$

$f( n +1)$ must be divisible by $n$

$f(4) \Rightarrow-6,-3,0,3,6$

$f(3) \Rightarrow-8,-6,-4,-2,0,2,4,6,8$

$f(2) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots, 8$

$f(1) \Rightarrow-8, \ldots \ldots \ldots \ldots \ldots, 8$

$\frac{f(4)}{3}$ must be odd since $f(3)$ should be even therefore $2$ solution possible.


	
		
			$f(4)$
			$f(3)$
			$f(2)$
			$f(1)$
		
		
			$-3$
			$2$
			$0$
			$1$
		
		
			$3$
			$0$
			$1$
			$0$

The number of functions $f :\{1,2,3,4\} \rightarrow\{ a \in Z :| a | \leq 8\}$ satisfying $f ( n )+$ $\frac{1}{ n } f ( n +1)=1, \forall n \in\{1,2,3\}$ is

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