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