By clue 2, the two numbers in the second row of Number Pyramid 7 sum
to 6, while by clue 5, the rightmost number in row 2 minus the number at the
apex of Number Pyramid 7 equals 1. Reducing these clues to equations,
letting the leftmost number in row 2 equal X, the rightmost number in row 2
equal Y, and the pyramid apex equal Z, we have X + Y = 6 for clue 2 and Z =
Y  1 for clue 5. Summing the two equations, we get X + Z = 5, meaning that
the leftmost number in row 2 and the apex of the pyramid must sum
to 5. By clue 3, the leftmost number in row 4 minus the 3rd number in row 4
equals 6. Therefore, there are four possible combinations for the leftmost
number in row 4 and the 3rd letter in row 4: 60, 71, 82, and 93. Trying
60, since the leftmost number in row 2 and the apex sum to 5, the leftmost
number in row 3 would be 9 (clue 1). However, there is no way for clue 4 to
work. Trying 71, since the leftmost number in row 2 and the apex sum to 5,
the leftmost number in row 3 would be 8 (1). However, since the 1 would
already be used, again there is no way for clue 4 to work. Trying 82, since
the leftmost number in row 2 and the apex sum to 5, the leftmost number in
row 3 would be 7 (1). However, since the 2 would already be used and
since 1 in the center of row 3 would make the rightmost number in row 3 8, there is
no way for clue 4 to work. Therefore, the leftmost number in row 4 and the
3rd letter in row 4 are 9 and 3 respectively. Since the leftmost number in
row 2 and the apex sum to 5, the leftmost number in row 3 is 6 (1). By
clue 4, the middle number in row 3 must be 1 or 2. If the middle number in
row 3 were 1, the rightmost number in row 3 would be 7. By clue 2 and the
numbers we would already have in the pyramid, the number combination for
row 2 would be 42 or 24. 42 cannot work because the apex would then be
1 (5) and 24 cannot work because the apex would be 3 (5). So, the middle
number in row 3 is 2 and the rightmost number in row 3 is 8. By clue 2 and
the numbers we would already have in the pyramid, the number combinations for
row 2 could be 51 or 15. If the combination were 15, the apex would be
4 (5). The two remaining numbers would be 0 and 7but this combination
cannot fit clue 6. So, row 2 has 5 and then 1 in it, with 0 at the apex (5).
By clue 6, the two remaining numbers of 4 and 7 must be the second and
rightmost numbers in row 4 respectively. In sum, Number Pyramid 7 is
filled as follows:


