Equal sums

By some reasoning we can find out that one digit must be 9. And 9 must be with 1 and 3.

And one digit must be 8.Then, 8 must go with either 1 and 4 or 2 and 3.

One digit must be 7, and it should be with either 1 & 5 or 2 & 4.

Also, one digit must be 6.And it should go with either 2 & 5 or 3 & 4.

Since, no digit may be used in more than 2 sums. The following cases arise:

Case 1: If 8 goes with 1 and 4, then 7 goes with 2 and 4 and so, 6 goes with 2 and 5.

Case 2: If 8 goes with 2 and 3, then 7 goes with 1 and 5 and so, 6 goes with 2 and 5.

But case 2 is impossible as the digit 4 does not occurs.

So, Case 1 is correct and from the diagram E must be 4.

These type of problems may be solved by trial and error, but the trial and error can be minimized by a proper use of logic.

This is a logic problem. There is a question with 4 different possible answers.Then,only one answer is right .You find one solution to the problem. This solution is the one only possible option between the answers. Then if the problem is right the solution must be your solution,Am I explaining myself?... Of course, I understand you, The solutions exposed here are not a proof,demostration of the result, but they are solutions for my comment before, It is logic. A proof: 9 can't be E because 9+3+1 is the only option with numbers 1, 2, 3, 4, 5, 6,7 ,8 ,9 containing 9 / its sum is 13, and E is in two different sums so 9 can´t be a solution for E and ...I dont have a proof, but I have answered 4,jaja, because I have found the solution of Dines and for logic...