The number of functions f : { 1 , 2 , 3 , 4 } { a : Z a 8 } satisfying f ( n ) + 1 n f ( n + 1 ) = 1 , n { 1 , 2 , 3 } is [2023]

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Solution

Q.
The number of functions $f : {1, 2, 3, 4} \to {a \in : Z | a | \leq 8}$ satisfying $f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$ is [2023]

1 2

2 1

3 4

4 3

Ans.
(1)

$f : {1, 2, 3, 4} \to {a \in Z : | a | \leq 8}$

$f (n) + \frac{1}{n} f (n + 1) = 1, \forall n \in {1, 2, 3}$

$f (n + 1)$ 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, ................, 8

$f (1) \Rightarrow$ -8, .............. , 8

$\frac{f (4)}{3} must be odd since f (3) is even.$

$∴$ Only two solutions possible.

$\begin{matrix} f (4) & f (3) & f (2) & f (1) \\ - 3 & 2 & 0 & 1 \\ 3 & 0 & 1 & 0 \end{matrix}$

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