Problem:
If x,y, and z are positive with xy=24,xz=48, and yz=72, then x+y+z is
Answer Choices:
A. 18
B. 19
C. 20
D. 22
E. 24
Solution:
Since
x=y24=48z
we have z=2y. So 72=2y2, which implies that y=6,x=4, and z=12. Hence x+y+z=22.
OR
Take the product of the equations to get xy⋅xz⋅yz=24⋅48⋅72. Thus
(xyz)2=23⋅3⋅24⋅3⋅23⋅32=210⋅34.
So (xyz)2=(25⋅32)2, and we have xyz=25⋅32. Therefore,
x=yzxyz=23⋅3225⋅32=4
From this it follows that y=6 and z=12, so the sum is 4+6+12=22.
Answer: D.
The problems on this page are the property of the MAA's American Mathematics Competitions