Find four positive integers A < B < C < D that satisfy both of these conditions:

(i) The geometric mean of A and D is 24, and:
(ii) The harmonic mean of A, B, C and D is 19.2

From (i) AD = 24^2 = 576, therefore the factors of AD are (2^6)*(3^2) and the possible pairs of A and D, respectively, are:
(1,576), (2,288), (3,192), (4,144), (6,96), (8,72), (9,64), (12,48), (16,36), and (18,32).

From (ii) 4/(1/A + 1/B + 1/C + 1/D) = 19.2
120/576 = (1/A + 1/B + 1/C + 1/D)

(A+D)/AD = (A+D)/576, thus A+D < 120, which eliminates
(1,576), (2,288), (3,192) and (4,144).

120-(A+D) = 576(B+C)/BC,  and given B and C are positive integers, the quadruplets {A,B,C,D) are
(8, 24, 36, 72) and (12, 16, 24, 28) as the two possible solutions.

Edited on December 16, 2014, 4:37 pm

Posted by Dej Mar on 2014-12-15 19:12:55