Question

A function f satisfies f(0)=0,f(2n)=n,f(2n+1)=f(n)+1 for all positive integers n.what is the value of f(2018).

Answer 1

F(2n) = n

Let 2018 be 2n then :

2n = 2018

n = 2018/2

n = 1009

Therefore :

f (2 × 1009) = 1009

As per the following fact :

f(2n) = n

Therefore :

f(2018) = 1009