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 Point to 2013 (2) (Posted on 2013-02-22)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 Outline of analytic solution Comment 2 of 2 |
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
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/hosted-sites/R.Knott/Fibonacci/cfINTRO.html
and if you want an app that does it for you:
http://www.maths.surrey.ac.uk/hosted-sites/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 2013-02-23 17:06:46

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