Function problems

Algebra Level 3

A function f f satisfies f ( 0 ) = 0 , f ( 2 n ) = f ( n ) f(0) = 0, f(2n) = f(n) , and f ( 2 n + 1 ) = f ( n ) + 1 f(2n + 1) = f(n) + 1 for all positive integers n n .

What is value of f ( 2017 ) f(2017) ?


The answer is 7.

Vijay Simha by Vijay Simha , 47, USA

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1 solution

Brian Moehring
Feb 12, 2017

Write 2017 2017 in binary as 1111110000 1 2 11111100001_2 . Then by the definition of f f , we have f ( 2017 ) = f ( 1111110000 1 2 ) = f ( 111111000 0 2 ) + 1 = f ( 11111100 0 2 ) + 1 = ⋯ = f ( 11111 1 2 ) + 1 = f ( 1111 1 2 ) + 1 + 1 = ⋯ = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7. f(2017) = f(11111100001_2) = f(1111110000_2) + 1 = f(111111000_2) + 1 = \cdots = f(111111_2) + 1 = f(11111_2) + 1 + 1 = \cdots = 1+1+1+1+1+1+1 = 7.

This doesn't merely compute the answer. It also tells us what f f actually is. In short, f ( n ) = digit sum of n in binary f(n) = \text{digit sum of } n \text{ in binary}

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