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(1/x)+(1/y)=(1/n) has 2005 ordered pairs as solutions to (x,y). I.e (y,x) is distinct from (x,y). The smallest n can be expressed as (x^y)(z^p). Where x and z are distinct primes. Find x+z+y+p.
The answer is 207.
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1 solution
Rearrange the equation to (x-n)(y-n)=n^2 Now note that n is a product of distinct primes y1,y2...yq that are repeated x1,x2,...xq times where xk>=1. n^2 has its primes repeated 2x1,2x2...2xq times. Total number of solutions is given by (2x1+1)(2x2+1)...(2xq+1)=2005=5(401). We see that q<=2 with q =2 we have x1=2 and x2=200 (x1 and x2 are arbitary). with q=1 we have x1=1002. The smallest solution with q=2 is (3^2)(2^200). With q =1 it is 2^1002. Clearly the former solution is better hence the solution is (3^2)(2^200)