Trollathon #2.6 : Mysterious Function
f ( 5 0 ) = 3
f ( 3 1 ) = 5
f ( 2 7 3 0 ) = 6
f ( 3 8 4 ) = 2
What is f ( 2 0 1 4 ) ?
The answer is 9.
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2 solutions
What's wrong with f ( x ) = − 1 9 0 0 6 8 6 3 8 5 2 8 4 1 6 8 0 2 0 2 7 8 7 4 7 7 3 x 3 + 1 2 6 7 1 2 4 2 5 6 8 5 6 1 1 2 4 2 9 9 6 8 1 2 7 9 3 9 x 2 − 9 5 0 3 4 3 1 9 2 6 4 2 0 8 4 0 1 2 5 6 4 8 6 0 5 2 8 9 1 9 1 1 x + 7 9 1 9 5 2 6 6 0 5 3 5 0 6 9 4 9 9 5 3 1 8 3 1 9 2 8 6 ? One can construct infinitely many f ( x ) using Lagrange Interpolation, thus giving infinitely many choices for f ( 2 0 1 4 ) . This is why I don't like these problems.
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This is a number theory problem, by the title, so the writer wants us to find a simple pattern, not use wolfram alpha.
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What I'm trying to say is that there are more than one possible answers. The question never stated f ( x ) cannot be a polynomial.
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@Sreejato Bhattacharya – who allows people to add such idiotic questions?
I agree with Sreejato. By newtons interpolation as we get many such functions, there is no uniquenss in the value of the question. Also when mathematics is talked , Sri.Phil Peters, the simplicity is not a matter of concern. Admit what Sreejato pointed out.
Expanding each term in base two, we find that the function counts the number of ones.
Basically f(x) counts the number of 1s in the binary representation of x. So, f ( 2 0 1 4 ) = 9 .