From the introduction, a different number 015 is in each cell of Number
Square 2. By clue 1, in the upperleft to lowerright 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 015 range of the cells is for one set of numbers to be 140 and
the other 151. First we test the sequence down the diagonal as 151014.
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
2nd4th rows of col. 1 (3415 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 24 possibilities down column 1 would be 865, 1054, and 1243. 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, 151054 down col. 1, fits. By clues 6 and 7,
then, row 4, col. 3 would hold 10 and row 4, col. 2 9but we would then have
two 10's in the square. Therefore, the 151014 diagonal cannot work; the
sequence down the upperleft to lowerright diagonal is 140115. 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 2nd4th rows of
col. 1 (3414 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 24 possibilities
down column 1 would be 965, 1154, and 1343. 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, 141154 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
8but 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
256712 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 015 as follows:
14 12 6 2
13 0 5 7
4 11 1 8
3 9 10 15


