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

enter image description here

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:

enter image description here

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.

+ 824

is an example.

  • $\begingroup$ The minimalistic regularity of the bottom row (sum) shows how much care this took to work out (spoiler: much!) $\endgroup$
    – humn
    Jul 24, 2017 at 3:57

2 Answers 2


The answer is:

Answer Image

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

  • $\begingroup$ "0 can't be added to any number because it will result in the same number" - not if there is a carry. $\endgroup$
    – ffao
    Jul 23, 2017 at 10:37
  • $\begingroup$ That would work in the addition itself, in the first and second column, you're right, but I meant it can't be added when I started thinking about the digits alone, not with the addition. I just posted how I thought about it and the steps I took to get to my answer, it's up to the OP to say if I went through it in the correct way, or it was just another lucky guess. $\endgroup$
    – Paul Karam
    Jul 23, 2017 at 10:40
  • $\begingroup$ It definitely reads to me that this is pretty much as much guesswork as mine. After about half way you just try things to see if they work and eliminate them if not. It isn't really what I'd call logical deductions. Still, upvote because your presentation of the answer was much prettier than mine. ;-) $\endgroup$
    – Chris
    Jul 26, 2017 at 8:40
  • $\begingroup$ I did try to follow some logical deduction, but I guess I've went through the wrong road too, waiting Oray to post the right way that we should have used. $\endgroup$
    – Paul Karam
    Jul 26, 2017 at 8:49

The answer is:


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

  • $\begingroup$ Only 346+752=1098 is a valid solution from the one you posted. Which I was going to post and saw your answer :p $\endgroup$
    – Paul Karam
    Jul 23, 2017 at 0:34
  • $\begingroup$ Oh yes. I reuse 1 rather a lot don't I? ;-) $\endgroup$
    – Chris
    Jul 23, 2017 at 0:36
  • 1
    $\begingroup$ I am not sure why you like it that much, but yeah ;) $\endgroup$
    – Paul Karam
    Jul 23, 2017 at 0:37
  • 1
    $\begingroup$ I allowed myself to further more edit it, explanation in the comment section, not sure if I only had to say it for you to edit, or is it okay if I did it myself. $\endgroup$
    – Paul Karam
    Jul 23, 2017 at 0:40
  • 2
    $\begingroup$ @Oray I know this is way old, but how about you post the original solution that should have been posted :) $\endgroup$
    – Paul Karam
    Dec 31, 2017 at 9:11

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