Problem:
Suppose that x,y, and z are three positive numbers that satisfy the equations xyz=1,x+z1​=5, and y+x1​=29. Then z+y1​=nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Notice that
5⋅29⋅nm​​=(x+z1​)(y+x1​)(z+y1​)=xyz+x+z1​+y+x1​+z+y1​+xyz1​=1+5+29+nm​+1​
Thus 144⋅nm​=36. so that nm​=41​ and m+n=5​.
OR
Because
5=x+z1​=x+xy=x+x(29−x1​)=30x−1
it follows that x=51​⋅y=24. and z=245​. Thus z+y1​=245​+241​=41​.
The problems on this page are the property of the MAA's American Mathematics Competitions