Problem:
Let @ denote the "averaged with" operation: a@b=2a+bβ. Which of the following distributive laws hold for all numbers x,y, and z?
I.II.III.βx@(y+z)=(x@y)+(x@z)x+(y@z)=(x+y)@(x+z)x@(y@z)=(x@y)@(x@z)β
Answer Choices:
A. I only
B. II only
C. III only
D. I and III only
E. II and III only
Solution:
If xξ =0, then I is false:
x@(y+z)=2x+(y+z)βξ =2x+y+x+zβ=2x+yβ+2x+zβ=(x@y)+(x@z).
On the other hand, II and III are true for all values of x,y and z :
x+(y@z)=x+2y+zβ=22x+y+zβ=2(x+y)+(x+z)β=(x+y)@(x+z)β,
and
x@(y@z)=2x+2y+zββ=2(22x+y+zβ)β=22x+yβ+2x+zββ=(x@y)@(x@z)β
The problems on this page are the property of the MAA's American Mathematics Competitions