From what I can collect, there are two options:

9143
3265
7314
1427

and

9143
2365
8214
1427

Be happy to test my math, however they each look legitimate to me.

This is my logic. Let me know if I am lacking one thing:

We’ve solely constructive integers to work with, so something in an equation that sums to five should strictly be lower than 5. Notably, B, F, J, and N are all in such an equation, so they have to all be lower than 5, which means they’re 1-4 in some order. Their sum should be 10, which lets us remedy for A utilizing equation 16: A = 9. Utilizing equations 2, 3, and 4, we now have B = 1, C = 4, and G = 6. Moreover, since B + F + J + N = 10 and F + J = 5, B and N should additionally sum to five, so we now have N = 4. Now, M = 1 by equation 10. Utilizing equation 11, D = 3.

At this level, equation 17 is our most restrictive, since we now have 4 numbers summing to 13 and two of them already sum to 10. Thus, Ok and O should be 1 and a couple of in some order. By equation 9, then, L should be 3 or 4, however it could’t be 3 as a result of D is already 3 (sudoku guidelines!), so it should be 4, making Ok 1 and O 2. This offers us P = 7 by equation 14 and H = 5 by equation 18.

Now we now have a sq. of numbers left, all of which fall beneath the jurisdiction of equations 5 via 8. (We are able to double test our work by utilizing equations 12, 13, 15, and 16, as plugging in our knowns makes these equal to five via 8.) As a result of E + F = F + J = 5 and E + I = I + J = 10, we all know that E should be equal to J, and moreover, I needs to be precisely 5 greater than F. From the very starting, we all know that J (and due to this fact E) should be lower than 5; 1 and 4 are dominated out from already being positioned in the identical row, however 2 and three each result in legitimate options, as proven above.