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Color Blind version:

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Here's a sudoku puzzle with a twist!

The numbers given in the squares are actually the sum of the numbers in that region. For example: 18= Sum of the numbers in the 4 yellow squares or 12= Sum of the numbers in the 2 green squares and so on.

So here are the rules:

  1. no row or column should have any digit more than once.
  2. Each 3x3 box should have each digit from 1-9 exactly once.
  3. The digits in the region should add up to the number given.
  4. No number is to be repeated in a region. For example: 10=6+4 and not 5+5.

I believe the rules are clear. So, go on solving it. The answer to be accepted should show most of the steps clearly(At least the first few steps, because they are a bit difficult).

  • 1
    $\begingroup$ Isn't this just killer sudoku? killersudokuonline.com $\endgroup$
    – Strawberry
    Jan 17, 2017 at 17:33
  • $\begingroup$ @Strawberry Yes, it is a killer sudoku. $\endgroup$
    – Sid
    Jan 17, 2017 at 17:35
  • $\begingroup$ @Strawberry not quite, Killer sudoku allows for duplicates in the same region as long as they don't violate the 3x3, row, or column rule. $\endgroup$
    – dcfyj
    Jan 17, 2017 at 19:36
  • $\begingroup$ I don't think you're allowed to say $10\neq5+5$. I'm pretty sure that violates mathematics... $\endgroup$
    – boboquack
    Jan 17, 2017 at 21:17
  • 1
    $\begingroup$ This could equally be considered a KenKen variant. $\endgroup$
    – Neil W
    Jan 17, 2017 at 22:52

1 Answer 1


The solution:


Thank you @Ian for making it more readable. I was quiet in a hurry...

As to start with the puzzle. We need to follow some rules to have this puzzle started:

- Each line, column and 3x3 box is worth 1, 2, .. 9 = 45 points.

That means we can have some numbers already filled in. Starting the puzzle on the box bottom left, we know:

The 17 value consists of 9 and 8. The bottom 5 and 6 pairs must be 1, 2, 3 and 5 and the 16 above the 17 consists of a 9 and 7.
When following the rule I just described, we know there is an amount of points missing in the bottom left box: 45 - 17 - 17 - 5 = 6. With the group of 5 having a 2 and 3, we know the remaining 6 points needs to be 1 and 5. Therefore, we put the 1 and 5 in both the group of 8 and the group of 6. We can now see the following:
First step

Step 2.

Let's continue with the 9 above the 16. As said, each box has 45 points, and we do know one number is left. 45 - 19 - 8 - 8 = 4. So, we can fill in both 4 and 5. We now know the 12 in the same box as well. 45 - 18 - 19 - 5 = 3.
Top center box contains a 3, that has to contain of a 1 and 2. We can now already assume what numbers are in the different boxes.
Step 2

Step 3.

Continuing on the top right box. The 23 has to be a 6, 8 and 9. We can also fill in the 5 and 2, as the 5 can only be in one place. Now the 2 is taken, we know the group of 5 consists of a 1 and 4. The 3 and 7 are left and placed below the 2 and 4. With this information we can fill in most of the top left box as well. Starting with the 7 (middle top) and 2 (same row). We can now fill in the box of 18. 18 - 2 - 7 leaves 9. The only possibility of this 9 is a 1 and 8.
Step 3

Step 4.

We continue with the middle left box. The 8 can only be placed within the group consisting of 9. We fill in both 8 and 1 here. Now the 3 and 5 has to go to the left of this group of 8, while the 2 and 6 are placed right. This allows us to complete the top left box as well.
I checked the middle box as well and I found out I can fill in one more free number. 45 - 16 - 11 - 13 = 5. This places the 5 on three new locations.
Step 4

Step 5. Are we getting the hang of it yet? ;)

Let's take a look at the 21 of the box right top. If I would fill in a 3, the puzzle cannot be solved, as I cannot fill in a 13. I have to add the 7 there, completing the top right box with a 3 in the 10 group. I placed a 7 below. In this row I now have left a 2, 3 and 6 which I am able to fill in the middle box. I tried putting a 2 in the group of 16 in the box, but I can't. 14 consists of 6 and 8, which is impossible to do. So the 2 goes in the box of 11. 3 and 6 makes 9, so we can add the 7 in the group of 16 as well.
Step 5

Step 6.

We are able to finish the middle box now.
Step 6

Step 7.

The 13 in the middle right box can only be a 3, 4 and 6. And, since the 3 and 4 are used in the last column, we know the exact location of the 6. I did exactly the same on the 19 in the bottom middle box.
Step 7

Step 8.

As said before, a 23 can only exist of a 6, 8 and 9. We are able to fill in the 23 in the bottom box now as the 8 must be placed in the box bottom right. With this information, we are able to add more numbers in the bottom box and in the middle right box.
Step 8

Step 9.

I assume the last part is the easiest one: bottom right box. The 16 of the middle box already has a 5, 1 and 3. 6 is missing there. The group of 8 can only have a 1, 3 and 4 combination. Let's fill that in, as well as the remaining box of 18. Now we are almost done - do not forget the 3 and 4 in the middle right box!
Step 9

  • $\begingroup$ Aside from explain how you came to this solution, could you make your solution more legible? At the moment it's a garbled mess. $\endgroup$
    – dcfyj
    Jan 17, 2017 at 16:57
  • $\begingroup$ I am very sorry to have you left this miss. I was in a hurry. Thanks to Ian, it's better readable now. I just added an explanation (in steps) as well. $\endgroup$ Jan 17, 2017 at 18:09

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