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Number Square 2
Putting the numbers 0-15 into a 4x4 square adds up to a Challenger Logic Puzzle.
June 14, 2004
From the introduction, a different number 0-15 is in each cell of Number Square 2. By clue 1, in the upper-left to lower-right diagonal, the number in the top left cell minus the number next down the diagonal equals 14, while the number in the bottom right cell of the diagonal minus the number next up from it also equals 14. The only way to get these results given the 0-15 range of the cells is for one set of numbers to be 14-0 and the other 15-1. First we test the sequence down the diagonal as 15-1-0-14. By clue 3, the number in col. 1, row 3 minus the value in col. 1, row 4 equals 1; while by clue 4, the numbers in the first column sum to 34. Since the only way to get the difference of 1 in clue 3 is to have an odd and an even number, which must sum to an odd number, given the remainder of 19 in the 2nd-4th rows of col. 1 (34-15 in the top cell), the number in col. 1, row 2 must be even. By clue 2, col. 2, row 3 minus col. 1, row 3 gives 7; therefore, the largest numbers that could be in col. 1, rows 3 and 4 are 6 and 5 (clue 3). From that and the fact that the number in col. 1, row 2 would be even, the smallest possible number in col. 1, row 2 would be 8. So, the rows 2-4 possibilities down column 1 would be 8-6-5, 10-5-4, and 12-4-3. The number in col. 2, row 3 for each of these arrangements then would be 13, 12, and 11, respectively. By clue 6, the sum across row 4 equals 37; while row 4, col. 3 minus row 4, col. 2 equals 1 (7). Clue 7 thus requires odd and even numbers in cols. 2 and 3 of row 4, in some order. Therefore, the difference between the 37 in clue 6 and the corner cells of row 4 must be odd. Only the middle arrangement above, 15-10-5-4 down col. 1, fits. By clues 6 and 7, then, row 4, col. 3 would hold 10 and row 4, col. 2 9--but we would then have two 10's in the square. Therefore, the 15-1-0-14 diagonal cannot work; the sequence down the upper-left to lower-right diagonal is 14-0-1-15. By clue 3, the number in col. 1, row 3 minus the value in col. 1, row 4 equals 1; while by clue 4, the numbers in the first column sum to 34. Since the only way to get the difference of 1 in clue 3 is to have an odd and an even number, which must sum to an odd number, given the remainder of 20 in the 2nd-4th rows of col. 1 (34-14 in the top cell), the number in col. 1, row 2 must be odd. By clue 2, col. 2, row 3 minus col. 1, row 3 gives 7; therefore, the largest numbers that could be in col. 1, rows 3 and 4 are 6 and 5 (clue 3). From that and the fact that the number in col. 1, row 2 must be odd, the smallest possible number in col. 1, row 2 would be 9. So, the rows 2-4 possibilities down column 1 would be 9-6-5, 11-5-4, and 13-4-3. The number in col. 2, row 3 for each of these arrangements then would be 13, 12, and 11, respectively. By clue 6, the sum across row 4 equals 37; while row 4, col. 3 minus row 4, col. 2 equals 1 (7). Clue 7 thus requires odd and even numbers in cols. 2 and 3 of row 4, in some order. Therefore, the difference between the 37 in clue 6 and the corner cells of row 4 must be odd. The second arrangement above, 14-11-5-4 down col. 1, does not work. From clues 6 and 7, then, for the first arrangement above, row 4, col. 3 would hold 9 and row 4, col. 2 8--but we would already have a 9 in col. 1. So, only the third order above works; and from clues 6 and 7, row 4, col. 3 contains 10 and row 4, col. 2 contains 9. By clue 5, an 8 fills cell row 3, col. 4. We now have numbers 2-5-6-7-12 unplaced. By clue 8, row 1, col. 3 minus row 1, col. 4 equals 4. The only possibility is for those cells to hold 6 and 2 respectively. By clue 9, row 1, col. 2 must contain 12 and row 2, col. 4 7. The 5 is in row 2, col. 3. in sum, Number Square 2 contains the digits 0-15 as follows:



       14 12  6  2
       13  0  5  7
        4 11  1  8
        3  9 10 15

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