Basic strategies (perhaps trivial but may be useful to someone). Fifth rule (more): Product of corners, border, inner area. And why the numbers in the corners of a 4x4 must be the same as those in the inner area (this property may be used in the timed puzzles).
Let’s call “Xb” the product of all numbers in the border (the “product value” of the border). Let’s call “Xc” the product of the four corners. Let’s call “Xi” the product of all numbers in the inner area (the “product value” of the inner area). Let’s call “p” the size of the puzzle (as we know the product of the numbers of a line, row or column, is a fix number, p!, “p factorial” (24 = 4 x 3 x 2 x 1, 120 = 5 x 4 x 3 x 2 x 1, 720 = 6 x 5 x 4 x 3 x 2 x 1, 5040 = 7 x 6 x 5 x 4 x 3 x 2 x 1, etc., for puzzles 4x4, 5x5, 6x6, 7x7, etc.).
If we multiply all the numbers in the top and bottom rows and in the leftmost and rightmost columns, that total will be (I use here the dot as the multiplication sign) p! . p! . p! . p! = (p!) ^ 4 (p factorial to the power of four) and it will be equal to the multiplication of the “border” by the four corners since we have multiplied the corners twice. So:
Xc . Xb = (p!) ^ 4 Xi . Xb = (p!) ^ p (this second equation because the border by the inner area constitute the full puzzle and the multiplication of all numbers in the puzzle equals p!.p!... (p times), that is, (p!) ^ p. Dividing these two equations:
Xi / Xc = (p!) ^ (p4) Xi = Xc . [(p!) ^ (p4)] or also Xc = Xi / [(p!) ^ (p4)] Now let’s see the special case of the 4x4’s: Xi = Xc. [(4!) ^ (4  4)] = Xc .[(4!) ^ 0] = Xc . 1 = Xc, the product of the corners is equal to the product of the numbers in the inner area (the square 2x2 inside). But not only this. Further, the numbers in the corners and in the inner area must be the same. We will verify this now using the product in a similar way as we did with the addtion. Let’s see all the possibilities:
Nrs in “heart” Xi Xc Factors Xc Corners (we eliminate combinations of the type 32=2224, for the corners)
1123 6 6 3211 1123 1124 8 8 2411 1124 1134 12 12 3221 3411 1134 (two 2’s in the corners force two 2’s in “heart” against the hypothesis) 2213 12 12 3221 3411 2213 (two 1’s in the corners force two 1’s in “heart” against the hypothesis) 2214 16 16 2241 4411 2214 (two 1’s and two 4’s in the corners force two 1’s and two 4’s in “heart” against the hypothesis) 2234 48 48 3224 3441 2234 (two 4’s in the corners force two 4’s in “heart” against the hypothesis) 3312 18 18 3321 3312 3314 36 36 3322 3341 3314 (two 2’s in the corners force two 2’s in “heart” against the hypothesis) 3324 72 72 3324 3324 4412 32 32 2441 4412 4413 48 48 3224 3441 4413 (two 2’s in the corners force two 2’s in “heart” against the hypothesis) 4423 96 96 3244 4423.
Now, three equal numbers inside the “heart” are impossible (two equal numbers contiguous). Finally, if the four numbers inside the inner area are different, 1234 (Xi = 24), the four numbers in the corners must also be 1234 (Xc = 24 has the only valid combination for the corners 3241), different numbers.
Looking to the general expression: Xi = Xc . [(p!) ^ (p4)], we observe that if “Xc” increases “Xi” also increases and vice versa; obviously if both increase simultaneously the “walls” in the border must decrease and vice versa. Let’s call “Xw” the product of all numbers in the four walls (the border without the cornes):
Xw = Xb / Xc = [(p!) ^ 4] / (Xc ^ 2) so if Xc increases Xw decreases and vice versa; in the case of a 6x6, for instance, Xw = (720 ^ 4) / (Xc ^ 2).
Let’s see this example (practical): The 6x6 difficult of June,15 (Puzzle Id: 305745) (we already used this puzzle for the topic “The use of maximums and minimums in the sum of cages”).
To start solving the puzzle the easiest way is obtaining C3C4, with a sum of 6 (the three leftmost columns  57): 24 is an impossible combination because if 4 is the higher number inside the cage “0“ then the pair in D3D4 should be 11, in other case since 1 is the minimum number possible in D3D4 the higher number in the cage would go to 7 (out of range). So the pair 15 is the solution (with 13 in D3D4, otherwise 17). In the other hand, considering the rows 1, 2 and 3, and since “1440x” has a sum of 22, C3D3 must sum 4 (63  59). The cross of the pairs 15 and 13 produces a 1 in C3 (and consequently in B1 and A2), etc. Once we have a sum of 6 in C4D4, considering the rows 4, 5 and 6, since “10x” is 125, the sum for “216x” is 18 (63  45), so 36621 is the only possible combination (being invalid the 33622, with a sum of 16, and 33641, with a sum of 17). In B6C6 we have a sum of 6 (in the “21+”) with the pair 24 since 5’s are not possible. The product of numbers in ”11+”, considering rows 2 and 3, is: X”11+” = 6!.6! / (2 x 180 x 3 x 12) = 720 x 720 / 12960 = 40, so the numbers inside the cage must be 524.
It is easy to see that the “10+” in B4B5C5, without 6’s or 5’s, has two possibilities 442 (impossible due to the presence of a 4 in B6C6) and 334 that becomes the correct one. But let’s determine that cage using only our formulas:
Xb = (X”9+”.X”1440x”.X”216x”.X”21+”) = 12 x 1440 x 216 x 960 = 3583180800 Xi = [(6!) ^ 6] / Xb = 720 x 720 x 720 x 720 x 720 x 720 / 3583180800 = 38880000
Now, X“10+” = 38880000 / (X”17+”.X”11+”.X”10x”.X”0“) = 38880000 / (180 x 40 x 10 x 15) = 36 And we have the combinations 334, 362 and 661 with the solution 334 (6’s are not possible).
Finally, Xc = [(6!) ^ 4] / Xb = 720 x 720 x 720 x 720 / 3583180800 = 75, so the only combination 5531 determines the corners: since 3 is in A1, 5 must be in F1, again 5 in A6 and 1 in F6. We knew almost from the beginning that the sum of the corners was 14 since C = I  42 = 56  42, and we could have followed other ways like the analysis of the “virtual triangle” in F1A6F6 (with a sum of 11, etc.), etc., but we have been just practicing.
