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