The impossible digital sum

Question

There are 10 digit numbers you are supposed to use shown as below;

And there is a very special addition where every digit is used only once. As you see, most of the digital signals (blue squares) are missing:

Can you fill the signals to form the right digits with the correct result?

Note: None of the numbers use a leading 0; so, neither 012 nor 0123, for example, are acceptable numbers.

  679
+ 824
------
 1503

is an example.

Answer 1

The answer is:

Reasoning I used to get it:

First, I followed the unfinished digits to put possible digits in their places, which gives us:

[2,3,5,6,8,9,0] [4,8,9,0] [4,5,6,8,9,0]
[1,2,3,4,5,6,7,8,9,0] [4,5,6,8,9,0] [1,2,3,4,5,6,7,8,9,0]
[1,2,3,4,7,8,9,0] [1,2,3,4,7,8,9,0] [1,2,3,4,7,8,9,0] [1,2,3,4,7,8,9,0]

Furthermore, we look at the hints, each digit is only used once, and we can't have a leading zero number, which eliminates the zero from multiple part:

[2,3,5,6,8,9] [4,8,9,0] [4,5,6,8,9]
[1,2,3,4,5,6,7,8,9] [4,5,6,8,9,0] [1,2,3,4,5,6,7,8,9]
[1,2,3,4,7,8,9] [1,2,3,4,7,8,9,0] [1,2,3,4,7,8,9,0] [1,2,3,4,7,8,9,0]
Zero at the end of first two rows is eliminated because it will surely give us a result with same digit twice, since any x+0 = x.

Now, since the result is a 4 digit number, and the summed numbers are both 3 digits number:

Even with same digits used twice such as 999+999 = 1998 (giving 999 is the max number we can combine using 3 digits), therefor the first digit in the result row is surely 1, which gives us the next:
[2,3,5,6,8,9] [4,8,9,0] [4,5,6,8,9]
[2,3,4,5,6,7,8,9] [4,5,6,8,9,0] [2,3,4,5,6,7,8,9]
1 [2,3,4,7,8,9,0] [2,3,4,7,8,9,0] [2,3,4,7,8,9,0]

Then I started looking for numbers that can be safely eliminated and I found:

In the result row, third column, we can easily eliminate the 8, because: Only way to get an 8 would be 8+0=8(invalid), 0+8=8(invalid) or 9+8 with carry on 1 = 8 (invalid). Not a big catch, but at least one less number.

Now what, I got stuck for a little bit here, or actually a lot. So I decided to try and find more numbers to eliminate, I found that:

Since the result of first column in first and second row must be > 9, in the second row if I wanted to use the two, it must be 8+2 = 10, and with no carry on 1. Therefor I went and tried it, it did give me a result such as
8 4 6
2 5 3
1 0 9 9
Which is invalid, so I can safely eliminate the 2 from first column, second row. Now I am left with:
[2,3,5,6,8,9] [4,8,9,0] [4,5,6,8,9]
[3,4,5,6,7,8,9] [4,5,6,8,9,0] [2,3,4,5,6,7,8,9]
1 [2,3,4,7,8,9,0] [2,3,4,7,9,0] [2,3,4,7,8,9,0]

Now I started thinking in another way since I felt that this was a roadblock because I couldn't find anything else to eliminate. I thought about the digits them self, 0 to 9. How can I use these 10 digits in additions to get unique results using each digit once.

0 can't be added to any number because it will result in the same number, which means using same digit twice, therefor 0 should be used in a result, not as an addition. What number can be found with this digits that has 0 in the result, only solution would mean the result is equal to 10. So now we have used 0 and 1 and the result should be = 10, so it's either 2+8, 3+7, 4+6, which led to me see that 9 can't be used in any more addition, so I need it as a result. Which can only be a 2+7, 3+6, 4+5. Now I have an 8 which is alone as well, so it must be a result as well. Which can only be the result of 2+6 and 3+5.
I felt that I am getting somewhere, so why not try and apply the rules I came up with to the addition we have. How to get an 8 first?
It can't be used in first column, since we need a result > 9, oh, that reminds me too that the only possibility is to have a result as 10 in first column, so 0 must be in the second column in the result row.

So now I had come to the conclusion where I have this:

[2,3,5,6,8,9] [4,8,9] [4,5,6,8,9]
[3,4,5,6,7,8,9] [4,5,6,8,9] [2,3,4,5,6,7,8,9]
1 0 [2,3,4,7,9] [2,3,4,7,8,9]
Now the only way to get an 8, is to have it in the last column of the result row, which leaves the 9 to be in the second column, therefor I am now down to this:
[2,3,5,6] 4 [5,6]
[3,4,5,6,7] [5,6] [2,3,5,6,7]
1 0 9 8

Wow, I am finally getting somewhere, now it's just easy calculations.

second column is surely now 4+5 = 9, which only leaves the 6+2 = 8, and 7+3 to the 10. So Final result would be equal to:
346 + 752 = 1098

Answer 2

The answer is:

346+752=1098

I worked out with a bit of luck. The logic I used was as follows:

Firstly if the bottom row is a four digit number it must start with a 1. We know from the rules it can't be a four digit number starting with a 0 and it is impossible to get any larger number in the first position by adding two three digit numbers (ie 999+999=1998)

Once this was set I thought about just looking at the information the puzzle gave us directly about numbers in positions:

(2/3/5/6/8/9) (4/8/9/0) (4/5/6/8/9)
(any but 1 or 0) (4/5/6/8/9/0) (Any but 1 or 0)
1 (2,3,4,7,8,9,0) (2,3,4,7,8,9,0) (2,3,4,7,8,9,0) (We have also eliminated 0 from the end of the first two numbers because it would trivially create a duplicate digit in the bottom row).

I then decided to make my life easier and try to find a solution where there was no more carrying than the one I already deduced. (this was the lucky guess).

This meant that the second column had to be 4+5=9.
Given that then the last column cannot have an 8 or 9 in the first two rows (if we did it would have to carry and we are looking for non-carry). This means the first row must be 6. The last row then must be 8 making the second row 2.

Once we have found places for 7 numbers it just remained to check if the last three were valid.

I was left with 3, 7 and 0 and fortunately 3+7 = 10 so we were done working out that 346+752 = 1098