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 xyâ‹…xzâ‹…yz=24â‹…48â‹…72. Thus
(xyz)2=23â‹…3â‹…24â‹…3â‹…23â‹…32=210â‹…34.
So (xyz)2=(25â‹…32)2, and we have xyz=25â‹…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