WELCOME TO 2015
Number Theory
Level
3
a,b are positive integers such that, 1/a+1/b=1/2015 how many solutions for a,b are there?? details and assumptions; .dont consider (b,a) if (a,b) exists
The answer is 13.
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1 solution
This can be factorised as :
(2015-a)(2015-b)= 2015^2
Number of ordered pairs satisfying the condition are 27 ,but, we don't want to consider (a,b) if (b,a ) already exists .
Thus, Number of solutions = (27-1)/2 =13 (since one solution is (4030,4030) )