Question 577782
Suppose the digits are a, a-d, a-2d. Then,


*[tex \LARGE A = a(2a-3d) + (a-d)(2a-2d) + (a-2d)(2a-d)]


*[tex \LARGE = 6a^2 - 12ad + 4d^2]


Additionally, *[tex \LARGE B = (a-d)(3a-3d) = 3a^2 - 6ad + 3d^2]. The problem tells us that A = (4/3)B so 


*[tex \LARGE 6a^2 - 12ad + 4d^2 = \frac{4}{3}(3a^2 - 6ad + 3d^2) \Rightarrow 2a^2 - 4ad = 0] (upon working out all the algebra)


This factors nicely to *[tex \LARGE 2a(a-2d) = 0]


Either a = 0 or a-2d = 0 (a = 2d). a cannot equal 0 because we have a three digit number (either the digits are negative or the number is 000) so a = 2d. If this is the case, then the only possible numbers are


a=2, d=1 --> 210
a=4, d=2 --> 420
a=6, d=3 --> 630
a=8, d=4 --> 840


4 possible numbers, answer is A.