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integer solutions of x+y+z=2013 so that xyz becomes max

Source: Greece JBMO TST 2012 p4

April 29, 2019
combinatoricspositive integersinteger solutions

Problem Statement

Numbers x,y,zx,y,z are positive integers and satisfy the equation x+y+z=2013x+y+z=2013. (E) a) Find the number of the triplets (x,y,z)(x,y,z) that are solutions of the equation (E). b) Find the number of the solutions of the equation (E) for which x=yx=y. c) Find the solution (x,y,z)(x,y,z) of the equation (E) for which the product xyzxyz becomes maximum.
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