Note: this is not in conjunction with my Minesweeper puzzles

Got this idea from one of Cracking the Cryptic's videos, I don't remember which one though. This exact puzzle is still overall unique.

Take this 4x4 grid below:


Now, you might not notice this, but this is a 4x4 sudoku, and there is a twist to it:

We can use the numbers 1-9

Now, you might be confused. Why can we? Well, that's because the grid has to satisfy these 18 conditions:

  1. Like a regular Sudoku, each number that is placed down can only be used once in each row and column.
  2. $A+B=10$
  3. $B+C=5$
  4. $C+G=10$
  5. $E+F=5$
  6. $E+I=10$
  7. $F+J=5$
  8. $I+J=10$
  9. $K+L=5$
  10. $M+N=5$
  11. $A+B+C+D=2A-1$
  12. $E+F+G+H=2A-2$
  13. $I+J+K+L=2A-3$
  14. $M+N+O+P=2A-4$
  15. $A+E+I+M=2A+2$
  16. $B+F+J+N=A+1$
  17. $C+G+K+O=2A-5$
  18. $D+H+L+P=2A+1$

Programming and brute-forcing is outright banned.

Difficulty: ★★★★★★


2 Answers 2


From what I can gather, there are two solutions:




Feel free to check my math, but they both look valid to me.

Here's my logic. Let me know if I'm missing something:

We have only positive integers to work with, so anything in an equation that sums to 5 must strictly be less than 5. Notably, B, F, J, and N are all in such an equation, so they must all be less than 5, meaning they are 1-4 in some order. Their sum must be 10, which lets us solve for A using equation 16: A = 9. Using equations 2, 3, and 4, we have B = 1, C = 4, and G = 6. Additionally, since B + F + J + N = 10 and F + J = 5, B and N must also sum to 5, so we have N = 4. Now, M = 1 by equation 10. Using equation 11, D = 3.

At this point, equation 17 is our most restrictive, since we have four numbers summing to 13 and two of them already sum to 10. Thus, K and O must be 1 and 2 in some order. By equation 9, then, L must be 3 or 4, but it can't be 3 because D is already 3 (sudoku rules!), so it must be 4, making K 1 and O 2. This gives us P = 7 by equation 14 and H = 5 by equation 18.

Now we have a square of numbers left, all of which fall under the jurisdiction of equations 5 through 8. (We can double check our work by using equations 12, 13, 15, and 16, as plugging in our knowns makes these equivalent to 5 through 8.) Because E + F = F + J = 5 and E + I = I + J = 10, we know that E must be equal to J, and additionally, I has to be exactly 5 more than F. From the very beginning, we know that J (and therefore E) must be less than 5; 1 and 4 are ruled out from already being placed in the same row, but 2 and 3 both lead to valid solutions, as shown above.

  • $\begingroup$ Yes, both of your solutions are valid (and was honestly not expecting a solution this early or for there to be 2 solutions) so good job on solving the puzzle! $\endgroup$
    – CrSb0001
    Oct 31 at 20:53

Since brute force and programming is not allowed, let us start thinking:

1) 16 variables and 17 linear equations plus one constraint, obviously mean that this problem is over-specified (meaning it has more information than necessary to solve it. Like asking you to draw a right triangle and giving you the length of all three sides.) So likely at least two of these 18 conditions are not necessary to solve this puzzle. Looking at Equations 6 and 8 we see that E=J, Same for 5 and 7. Thus we can leave out any one of the equations 5-8 and still solve the puzzle.

- Eq 13: Let us use the substitution principle, i. e. substitute 10 for I+J (Eq. 8) and 5 for K+L (Eq. 9), that results in 15 = 2A-3 or A=9
- Eq. 2 then means B=1
- Eq. 3 then means C=4
- Eq. 4 then means G=6
- Eq.11 then reads 9+1+4+D=17 or D= 3
- Eq.12 then reads (with Eq. 5: E+F=5) 5+6+H=16 or H= 5
- Eq.15 then reads (with Eq. 7: E+I=10) 9+10+M=20 or M= 1
- Eq.16 then reads (with Eq. 5: E+F=5) 1+5+N=10 or N= 4 (This also means equation 10 is not needed)

At this point it makes sense to take stock of what we know and use it in the remaining equations: remember E=J:
- 5. E+F= 5
- 6. E+I=10
- 9. K+L=5
- 14. 5+O+P=14 thus O+P=9
- 17. 10+K+O=13 thus K+O=3
- 18. 8+L+P=19 thus L+P=11 By adding equations 17 and 18, we see that one of the Equations 9, 14, 17, 18 is redundant and not needed.
- From Equation 17 we conclude K=1 and O=2 or K=2 and O=1. Since O=1 is not possible because of M=1 in the same row, the only Solution is K=1 thus O=2, L=4 and P=7. From equation 5 we conclude that E can only be 1,2,3 or 4. 1 is not possible because of M=1 in the same column. E=4 is impossible since J=E=4 and L=4 are in the same row.

This leaves the two solutions E=J=2 , F=3, I=7 and E=J=3, F=2, I=8. It is easy to check that both solutions fulfill the condition that each number is only used once in each row and column.

A better (more challenging with a unique solution) puzzle would be achieved by omitting equations 8,9 and 10 and adding the equation J+L=6. This would make the solution non-trivial by preventing to easily conclude A=9 from equation 13, since neither (I+J) nor (K+L) are given.


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