Function Sums to One

Level pending

A function f : { 1 , 2 , , 2014 } R + f:\{1, 2, \cdots , 2014\} \rightarrow \mathbb{R^+} satisfies the following relation.

i = 1 ; i j 2014 f ( i ) f ( j ) = 1 j { 1 , 2 , , 2014 } \sum \limits_{i=1; \ i \neq j}^{2014} f(i) f(j) = 1 \quad \forall \ j \in \{1, 2, \cdots , 2014 \} The sum of all possible values of f ( 2014 ) f(2014) is k k . Find 1 k 2 \dfrac{1}{k^2} .

Details and assumptions

  • The range of f f is R + , \mathbb{R^+}, i.e. f ( x ) f(x) is always positive.

  • If no such functions exist, k = 0 k=0 .

  • If f ( 2014 ) f(2014) can take only one value, k k equals the only possible value of f ( 2014 ) f(2014) .

This problem appeared in the Proofathon Algebra contest, and was posed by Paramjit Singh.


The answer is 2013.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Cody Johnson
Feb 21, 2014

f ( x ) = ± 1 2013 f(x)=\pm\frac1{\sqrt{2013}} so the answer is bad.

f ( x ) > 0 f(x)>0

Adrian Neacșu - 5 years, 8 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...