a = 2

Using again the relation of order between b, c and d I found b<6. Because b>a, I have to analyse b=3, b=4 and b=5.

b = 3

From the relation between c and d, and substituting the values for a and b I obtained c<12, it means it could be 4, 5, 6, 7, 8, 9, 10 or 11.

c = 4


1 2 + 1 3 + 1 4 + 1 d =1 has no solution for d integer.

c = 5


1 2 + 1 3 + 1 5 + 1 d =1 has no solution for d integer.

c = 6


1 2 + 1 3 + 1 6 + 1 d =1 has no solution for d integer.

c = 7


1 2 + 1 3 + 1 7 + 1 d =1 gives d=42 .

c = 8


1 2 + 1 3 + 1 8 + 1 d =1 gives d=24 .

c = 9


1 2 + 1 3 + 1 9 + 1 d =1 gives d=18 .

c = 10


1 2 + 1 3 + 1 10 + 1 d =1 gives d=15 .

c = 11


1 2 + 1 3 + 1 11 + 1 d =1 has no solution for d integer.

b = 4

Repeating the same procedure, I obtained: c<8, and because c>b, c could be 5, 6 and 7.

c = 5


1 2 + 1 4 + 1 5 + 1 d =1 gives d=20 .

c = 6


1 2 + 1 4 + 1 6 + 1 d =1 gives d=12 .

c = 7


1 2 + 1 4 + 1 7 + 1 d =1 hasn't a solution for d integer.

b = 5

I obtained: 3c<20, and consequently only c=6 is possible. This hasn't a solution for d.

a = 3

For b, I obtained: 2b<9, and because b>a, it is possible only b=4. Then, I obtained 5c<24, i. e. without solution because c>b.

Finally, the good solutions are:

a 2 2 2 2 2 2
b 3 3 3 3 4 4
c 7 8 9 10 5 6
d 42 24 18 15 20 12