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6 x 6 Sudoku puzzle subdivided into 2 x 3 sub-grids

Normal Sudoku rules for this grid size apply. The product of the numbers on each pink diagonal equals the sum of the numbers in the blue square.


This puzzle is from the now closed contest at: https://rss.org.uk/RSS/media/File-library/Quiz/RSS-Christmas-Quiz-2023-R2.pdf

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  • $\begingroup$ Is there supposed to be a fairly short and straightforward solution path to this?? $\endgroup$
    – PDT
    Mar 16 at 16:01
  • $\begingroup$ Up to a certain point It seems you are just brute forcing values until one fits. $\endgroup$
    – PDT
    Mar 16 at 16:03
  • $\begingroup$ @PDT I am not sure if there is a simple solution to this puzzle. I decided to post this puzzle because it seemed elegant to me. $\endgroup$ Mar 16 at 19:57

2 Answers 2

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The magic number is:

30

Solution:

enter image description here

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  • $\begingroup$ I got the answer too but I can’t seem to find a simple enough solution path that what is difficult $\endgroup$
    – PDT
    Mar 16 at 14:18
  • 1
    $\begingroup$ Answers to [grid-deduction] puzzles on this site are generally expected to share at least part of their solve path. Can you edit in some key deductions? For example, how you determined the magic number. $\endgroup$
    – bobble
    Mar 16 at 14:19
  • $\begingroup$ @Bobble this puzzle is really difficult the only way I can see solving it is after finding out some possible values for the diagonals inferring values on the bottom three blue squares and then just trying out different permutations, see their refutations until one fits… $\endgroup$
    – PDT
    Mar 16 at 14:22
  • $\begingroup$ The diagonals can only be 30,32 and 36 in size the 32 because of the 442 configuration is easy to refute but the 36 configurations has a lot of different variations that you have to refute. Then you are left with the 30 and again there are a ton of refutations!!! $\endgroup$
    – PDT
    Mar 16 at 14:25
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    $\begingroup$ Some grid deductions are just not meant for PSE or where one is expected to give a thorough and concise step by step solution path… $\endgroup$
    – PDT
    Mar 16 at 14:27
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This is a devious sudoku despite its size, so it does deserve a step-by-step explanation. I am hereby writing down a solve-path.

Let's start by identifying the sum of the digits in blue cells.

The upper two-thirds of the box contain the digits 1-6. The other 3 cells can carry a sum between 1+2+3=6 and 4+5+6=15. Therefore the sum ranges between 27 and 36.

Now consider the prime factorizations of these values.

27=33
28=22×7
29 (prime)
30=2×3×5
31 (prime)
32=25
33=3×11
34=2×17
35=5×7
36=22×33

The numbers whose prime factorization includes primes greater than 6 are eliminated first.
Then looking at the shorter pink diagonals, we can say that the sum should be able to be written as a product of three different digits, to not violate the sudoku rules. Therefore 27 and 32 are excluded (32 can be written as 3×3×4, which is the only way it can be expressed as a product of 3 digits less than 6).

Then we are left with 30 and 36.

If it is 36,

the other digits must be 4,5 and 6. Since 5 is not a factor of 36, it shouldn't be in any pink diagonal. Then we get
1

Note that the circled cells can only be 4 or 6. However it can neither be (4,4), (4,6) nor (6,6), because the first two pairs disagree with the product and the last one forces every other cells along the long diagonal to contain 1s #. Therefore 36 is also discarded.

Then the sum is 30.

30= 2×3×5
30= 5×6
To get that sum the other three digits must sum to 9, giving three possibilities: 9=4+3+2
9=5+3+1
9=6+2+1

(2,3,4)

Similar to the logic used for 36, 4 cannot be in the product, so neither can be in the diagonals. Then we can pencil-mark these values.

2

But note that both possibilities introduce higher powers of 2 or 3 which leads to contradiction.

3

Therefore it cannot be (2,3,4).

(1,3,5)

4
If R3C1 isn't 5, then short / diagonal should contain a 5 in one of the circled cells, and short \ diagonal should contain a 5 in one of the squared cells. In the same time, 5 is in either R3C2 or R3C3 considering the middle left box. But we cannot have three 5's in 2 columns. Therefore R3C1 must be 5.*

5
Still, there should be a 5 in either of the marked squares, therefore 5 in the long / diagonal should be in one of the triangles. Then if R3C5 is 6, two 1s should go in the bottom left box violating sudoku rules, therefore it must be 2. Then there should be a (1,3) pair in the tail of the long / diagonal.

6
Now see that there cannot be 1 and 3 in the circled cells, so we cannot make our product equal to 30 in the short / diagonal.

After all these, we can conclude that

it is neither (1,3,5).

Thus, it is

(1,2,6)

Now we can leisurely solve the sudoku.

7

As explained in * paragragh above, similarly we can deduce that R4C1 is 5.

And if R4C4 is 6, then R3C5 is 5, and every other cell along the long / diagonal is 1; contradiction.

Similarly, if R3C3 is 6, then 5 is in either top left box or bottom right box, and at least one box gets two 1's along the long \ diagonal; contradiction.

8

5 in the bottom left box can be in either R6C2 or R5C3. Therefore there cannot be a 5 in R3C5 considering the long / diagonal; it should contain 3.

Pencil-marking some more, we get to this:

9

The next step is not so obvious (well, for me). So we will have to do some back-tracking. Assume that R3C1 is 1.

10

So it's not 1 :( (Note that all of the above are forced.)

If it is 6, then there should be a 1 (R1C3 or R2C2) and (R3C2 or R3C3) and (R5C2-R6C3 region): contradiction (recall the * paragraph).

Therefore R3C1 is 2.

Continuing,

11

(Actually I am stuck here as well. It's backtracking time!)

Assume R2C2 is 3.

12

Then it must be 5.

13

Done.

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  • $\begingroup$ Haha I remembered this tough sob. I remembered just going through a gargantuan amount of refutations to find the answer. It was like they chose the hardest possibility from the lot! It seems there is no shortcut from reading your answer. $\endgroup$
    – PDT
    May 11 at 17:52
  • $\begingroup$ It took me several hours to solve! Nice answer +1 $\endgroup$
    – PDT
    May 11 at 17:56

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