Just for fun!-XI

Algebra Level 3

Let F ( x ) F(x) be a unique polynomial of degree 4 4 such as F ( 165 ) = 20 F(165)=20

F ( 42 ) = F ( 69 ) = F ( 96 ) = F ( 123 ) = 13 F(42)=F(69)=F(96)=F(123)=13

Compute F ( 1 ) − F ( 2 ) + F ( 3 ) − F ( 4 ) + . . . + F ( 165 ) F(1)-F(2)+F(3)-F(4)+...+F(165)


The answer is 20.

Swapnil Das by Swapnil Das , 20, India

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1 solution

Rimson Junio
Aug 18, 2015

Let F ( x ) = k ( x − 42 ) ( x − 69 ) ( x − 96 ) ( x − 123 ) + 13 F(x)=k(x-42)(x-69)(x-96)(x-123)+13 . Notice the symmetry of the graph with respect to x = 82.5 x=82.5 such that F ( 1 ) = F ( 164 ) , F ( 2 ) = F ( 163 ) , F ( 3 ) = F ( 162 ) , . . . F(1)=F(164), F(2)=F(163), F(3)=F(162), ... . Then F ( 1 ) − F ( 2 ) + F ( 3 ) − F ( 4 ) + . . . + F ( 165 ) = F ( 165 ) = 20 F(1)-F(2)+F(3)-F(4)+...+F(165)=F(165)=20

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