Question

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

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

Answer: (A)

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