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Killer-X-tra (Posted on 2008-05-27) Difficulty: 3 of 5

Place the numbers 1 to 9 once in each row, column, long diagonal (marked with a heavy dot) and every 3x3 grid.

The 'cage' inclusions (individually coloured regions) indicate the sum or the product of the numbers in those cells.

Unlike Killer Sudoku, the same number may appear more than once in a cage. So, 36x in a cage which overlaps the 3x3 grids may contain two 6s and a 1.


My inspiration for this puzzle came from a variety of puzzles posted by Pete and Will at http://sudexel.com/forum/index.php

My thanks goes to brianjn for his hard work in producing the graphic.

No Solution Yet Submitted by Josie Faulkner    
Rating: 3.5000 (2 votes)

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Hints/Tips re(2): Solution | Comment 13 of 19 |
(In reply to re: Solution by Dej Mar)

The spreadsheet was only a grid template of which I could copy-paste to test out possible options, as well as, to provide possible sets of digits for the various calculations.

For those who wish to solve the Killer-X-tra using logic and would like to have much of the larger grid filled, here is a beginning:

For each cage with a given product, we need only examine the combination of factors such that no factor exceeds 9 so to have only numbers of single digits.

For 80x, we have the two possible sets of three numbers: (2,5,8) and (4,4,5). Due to the fact that its cage is in a single row and each row can only contain one of each distinct digit, we can exclude the set (4,4,5), leaving the set of three numbers (2,5,8),

For 28x, we have two sets of three numbers: (1,4,7) and (2,2,7). Due to the limitation of distinct digits for each 3x3 grid, for which the 3-cell cage entirely belongs, we can exclude the set (2,2,7).

For 56x, we have only one set of three numbers: (2,4,7). We know from the 80x cage in the first row that the 2 of the 56x cage must be in the second row.

For 4+, there are two sets of two numbers of single digits: (1,3) and (2,2). With the rule for distinct digits per column, we can exclude the set (2,2).  And, as there is already a 1 in the top-center 3x3 grid for the 28x cage, the 4+ cageís top cell must be a 3.

For 12+, there are four sets of two numbers of single digits: (3,9), (4,8), (5,7), and (6,6). With the rule for distinct digits per column, we can exclude the set (6,6).  As we already have a 4 and 7 for the 28x cage and a 5 and 8 from the 80x cage for the 3x3 grid, we can also eliminate the two sets (4,8) and (5,7), leaving only (3,9).  And, as we already have a 3 in the top-center 3x3 grid given by the 4+ cage, our top cell for the 12+ cage must be a 9.

As the 80x and 28x cages contain both 4 and 7, we know that one and only one of the two numbers can in the shared row in each cage. With a 1 in the 6 column in the 4+ cage, we know the 28x cageís cell in the sixth column cannot contain a 1, this limits the 1 of the 28x cage set to be in the center cell of the 3x3 grid. And, as we know each 3x3 grid contains one each of the nine digits, we know the center-left cell of the same 3x3 must contain the digit 6

For 336x, we have only one set of three numbers: (6,7,8). As there already exists a 7 in the top-right 3x3 grid in the 56x cage, the 7 must be in the lower cell of the 336x cage.  The 7 in the eighth column removes the possiblity that a 7 in the 56x cage is in the 3x3 gridís center cell.

For 40x, we have only one set of two numbers: (5,8). As the 40x cage entirely lies in the fourth column, we can deduce the 80x cage cell in the fourth column must then be a 2.

For 20x, we have only one set of two numbers: (4,5).  As the 20x cage entirely lies in the sixth column, we can deduce the 80x cage cell in the sixth column must then be an 8, and subsequently the middle cell of the 80x cage is then a 5. We also know, due to the digits in the 20x cage, that the middle-right cell of the middle-top 3x3 grid must be a 7, and hence the bottom-middle cell of the middle-top 3x3 grid must be a 4. From this we deduce that the top-right cell of the top-right 3x3 grid is a 7.

For 112x, we have two sets of three numbers: (2,7,8) and (4,4,7).  As there now already exists a 7 in the third row, we can eliminate the possibility of the (4,4,7) set as two 4ís also cannot share the same row.

For 252x, we have two sets of three numbers: (4,7,9) and (6,6,7). With the restriction of distinct digits in a column, we can exclude the (6,6,7) set for the yellow 252x cage.  Also, for the same restriction, we can exclude the (6,6,7) set for the blue 252x cage, as a 6 already exists in either the seventh or eigth column from the 336x cage set.  As a 7 already exists in the eight column  from the 336x cage set, the 7 in the blue 252x cage set must be in the sixth column.

For 27x, we have only one set of three numbers: (1,3,9).  As a 1 already exists for the bottom-left to top-right diagonal, we know the 1 for the 27x cage must be in the second column of the larger grid.

For 540x, we have three sets of four numbers: (2,5,6,9), (3,4,5,9) and (3,5,6,6).

With continued deducing and grid-filling, one will reach the same solution as given by Penny.


  Posted by Dej Mar on 2008-05-29 02:31:53
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