The “power” of the corners: full analysis 6x6 2012-Apr-14 (this post is dedicated to beaker, jomapil, … , and all those calcudokers who like the “addition rule” so that they may start liking the “multiplication rule” as well).
When we initially deal with a puzzle like this (6x6 difficult, Apr 14, 2012, Puzzle id: 446084):
we are inclined to think of a great complexity that in fact does not exist as we will see inmediately. The puzzle itself (with an assigned solving rate of 48.7) is not “officially” considered “very” difficult (and in fact it was solved by many puzzlers). But, in my opinion, it is a very interesting calcudoku for two reasons: 1) it can be solved applying only “analytical means” and 2) because of the possibility of discussing the relevance of the value of the corners (useful in some specific puzzles) and incidentally the content of the “inner area”, etc., apart of practicing with the general “multiplication rule”.
Perhaps we could have started breaking into prime factors all products involved (knowing that the products in the border cann’t repeat a number more than twice). Though we will not need it this time (because, along the process, we will not use at any moment any sum or the “addition” rule), I have also included (in parentheses) the “addition value” for every combination:
“96x” = [1,1,4,4,6] (16), [1,2,2,4,6] (15), [1,2,3,4,4] (14); “900x” = [1,5,5,6,6] (23), [2,3,5,5,6] (21), [3,3,4,5,5] (20); “12x” = [1,2,6] (9), [1,3,4] (8), [2,2,3] (7); “80x” = [1,4,4,5] (14), [2,2,4,5] (13); “540x” = [1,3,5,6,6] (21), [2,3,3,5,6] (19); “4x” = [1,1,4] (6), [1,2,2] (5); “288x” = [1,2,4,6,6] (19), [1,3,4,4,6] (18), [2,2,3,4,6] (17), [2,3,3,4,4] (16).
In a quick look we see that, since in c3 or c4 we can place (at most) a single 5 and there is another 5 in the cage “540x”, to complete the necessary three 5’s in the three leftmost columns we need a third 5 inside “16+” but, in these conditions, to complete the cage “16+”, the rest is 11, we need 5 + 6 (unique) so “16+” = [5,5,6]. This has two consequences: a6 = 5 (the only place for the 5 of “540x”) and d4 = 5 (it must be a 5 in the cage “80x”).
Comment: Supposing that the puzzler does not take notice of those 5’s at a first glance there is always a method to quickly advance to the solution: applying the multiplication property (though in this case managing huge products). The multiplication of the numbers of any line in a 6x6 is 6 x 5 x 4 x 3 x 2 x 1 = 720 (named 6!, factorial of 6). The multiplication of the three numbers inside the cage “16+” by the three numbers inside the cage “2:” must be equal to:
[(720) ^ 6] / (96 x 900 x 12 x 80 x 540 x 4 x 288) = 2700,
that is, the multiplication of all the numbers in the full grid (720 x 720 x … x 720), six lines, divided by the seven known products. Now, breaking into prime factors this result 2700 = 2 x 2 x 3 x 3 x 3 x 5 x 5. No 5 can go to the cage “2:” so both 5’s must go to the cage “16+” (numbers in blue) with a 6 (which gives the difference to 16). In these conditions the numbers inside “2:” must a have a product of 2700 / (5 x 5 x 6) = 2700 / 150 = 18, that is, “2:” = [1,3,6], [2,3,3,]. Obviously the only solution for “2:” is [1,3,6].
Next: The product of the cells c4 and d4 (looking to the three bottom rows) is:
c4 x d4 = (720 x 720 x 720) / (540 x 150 x 4 x 288) = 4 >>> c4-d4 = [1,4].
Consequently, c3 = 80 / (5 x 4) = 4 and it follows: c4 = 1, d4 = 4. Now the combination [1,1,4] is not possible for “4x” so ”4x” = [1,2,2] >>> e4 = 2, d5 = 2 and e5 = 1. And then d2 = 1 being the pair e2-e3 = [3,6].
Also (as explained in the above comment) the product of the numbers in the cage “2:” equals to:
(720 x 720 x 720) / (96 x 900 x 12 x 20) = 18.
And, since [2,3,3] is not valid, “2:” = [1,3,6] with d2 = 1 and e2-e3 = [3,6].
Next graphic:
Now we are going to define the other three corners since we can calculate the product of the corners. We must remember (see my post “Why in a 4x4 numbers in corners are those in inner area (2)” in the section “Solving strategies and tips”) that the product of the corners, named Xc, multiplied by the product of the numbers in the border, named Xb, equals (in a 6x6) to (6!) ^ 4 = 720 x 720 x 720 x 720. The reason for this is that if we multiply all numbers in the exterior rows, 1 and 6, and all numbers in the exterior columns, a and f, this total product is equal to 720 ^ 4, having multiplied all numbers in the border but including twice the corners so we have the product of the corners by the product of the border itself, Xc by Xb. Then, in our puzzle:
Xc = (720 x 720 x 720 x 720) / (96 x 900 x 288 x 540) = 20.
The product of the corners is 20 and being a6 = 5 we are left with two possibilities: (a1 =1; f1 = 4; f6 = 1) or (a1 = 2; f1 = 1; f6 = 2). If f1 = 4 we are dealing with the combination [3,3,4,5,5] for the cage “900x”(the position of the numbers is shown in red colour in the above graphic) and it can be quickly observed that now it would not be possible to place a 2 in column f (f4 or f5). As a consequence (in green): a1 = 2, f1 = 1, f6 = 2. The corners are now defined, they “inform” the full solution, and we can affirm that the puzzle is solved.
A final comment: Though it is not really necessary to calculate the solution for the cage “12x” (in this process it will be derived soon and straight forward) it’s interesting to underline now that this cage must be [2,2,3] for three reasons:
a) “96x” contains a 1 (and only once, due to f1 = 1 and d2 = 1, so [1,1,4,4,6] would never be possible); no more 1’s can go in any other place of the three upmost rows (or, inversely, if “12x” is [1,2,6] or [1,3,4] >>> b3 = 1 and there is no place for a 1 in the other positions of the cage “96x” but any of the possible combinations for this cage require a 1).
b) Since we have two 2’s in the corners, a total of four 2’s must go to the “inner area” of the puzzle (this is explained in the thread “Why in a 4x4 numbers in corners are those in inner area (1)”, in the same section, where it is shown that, in a 6x6, the 16 numbers in the “inner area”, that is the 4x4 inner box, are twice the set 1, 2, …, 6 plus the four corners) so both 2’s must go to the cage “12x”.
c) The “addition value” of “the inner area” must be 10 + 42 = 52, the addition of the corners plus 2 lines (2 x 21) (see referred posts), then “12x” (+) = 52 - “2:” (+) - “80x” (+) - “16+” (+) - “4x” (+) = 52 - 10 - 14 - 16 - 5 = 7 >>> “12x” = [2,2,3].
We complete the puzzle without any additional calculation. Logically if f1 = 1, the cage “900x” = [1,5,5,6,6]. The following two graphics are self-explanatory:
The “official” solution:
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