Q2014/004) p(x) = a2012 x^2012 + a2011 x^2011 + … + a1 x + a0 a2012 etc., are just coefficients. and p(x) = 1/x for integer x = 1,2,3,…,2013 What is p(2014)

We have xp(x) – 1 = 0 at 1,2,…. 2013 and of order 2013

so xp(x) = 1 – m(x- 1) ..(x-2013)
at x =0 we get 1 – m(-1)(-2)(-2013) or 1 + m * 2013! = 0
m = -1/2013!
at x = 2014
2014p(2014) = 1 + 1/2013!(2013) !
= 1 + 1 = 2
or p(2014) = 2/2014 = 1/1007