Each of p and q is a
positive integer with gcd(p, q)=1.
Determine the minimum value of p+q such that:
 The first four digits immediately following the decimal point in the base ten expansion of p/q + q/p is 2013.
This is probably a lot more work than brute force (especially with a computer) but it does outline a way to do it analytically (or even by hand).
Let n be the integer part of the number sought. The smallest possible sum of p/q+q/p is 2 so the smallest possible n=2. So we begin here:
Find an approximate solution to
p/q+q/p = 2.20135
Rewrite as
p^2  2.21035pq + q^2 = 0
The quadratic formula gives
p = (2.20135q + √((2.20135q)^2  4q^2))/2
or
p/q = (2.20135 + √((2.20135)^2  4))/2
[note if you use the  instead of the + you just get the reciprocal. p and q are just interchanged]
p/q ≈1.560550478...
Now to find a small rational equivalent this site explains how to do it by hand
http://www.maths.surrey.ac.uk/hostedsites/R.Knott/Fibonacci/cfINTRO.html
and if you want an app that does it for you:
http://www.maths.surrey.ac.uk/hostedsites/R.Knott/Fibonacci/cfCALC.html#
The first few rational approximations are not close enough. The first that is close enough is 66/103
for a p+q = 169
In retrospect it starts to become brute force here:
To show whether of not this the smallest you could next use n=3 and so on (I stopped here.) The procedure does not need continue indefinitely since as n increases so does the value that p/q must approximate. In fact it converges to n + .2013

Posted by Jer
on 20130223 17:06:46 