| By clue 4, the rightmost number in the 2nd row of Number Pyramid 12
minus the leftmost number in row 2 equals 6, so that the possible
left-to-right combinations are 0-6, 1-7, 2-8, and 3-9. If the numbers in
row 2 were 0 and 6, since the apex and the rightmost numbers in rows 2, 3,
and 4 sum to 26 (clue 2), the numbers in the apex and rightmost in rows
3 and 4 would be 9-8-3, 9-7-4, or 8-7-5 in some order. Since the rightmost
number in row 2 is larger than the apex number (4), the 3, 4, or 5
respectively would have to be at the apex. However, neither the 9 nor 8
can be in row 3, since the whole row sums to 8 (5); nor could the 7 of
9-7-4 or 8-7-5 be in row 3, since the other two numbers would be
1 and 0--which would already be used. So, the first combination of 0 and 6
across row 2 cannot work. If the numbers in row 2 were 1 and 7, since the
apex and the rightmost numbers in rows 2, 3, and 4 sum to 26 (2), the numbers
in the apex and rightmost in rows 3 and 4 would be 9-8-2, 9-6-4, or 8-6-5
in some order. In the 9-8-2 case, since the rightmost number in row 2 is
larger than the apex number (4), the 2 would have to be at the apex. However,
neither the 9 nor 8 can be in row 3, since the whole row sums to 8 (5). In
the 9-6-4 case, since the rightmost number in row 2 is larger than the apex
number (4), either 6 or 4 would have to be at the apex. If 6 were at the
apex, since 9 can't be in row 3 (5), 4 would be rightmost in row 3. However,
by clue 5, the other two numbers in row 3 would have to be 3 and 1--impossible,
since the 1 would already be used. If 4 were at the apex, since 9 can't be in
row 3 (5), 6 would be rightmost in row 3. However, by clue 5, the other two
numbers in row 3 would have to be 0 and 2--contradicting clue 1. In the 8-6-5
case, since the rightmost number in row 2 is larger than the apex number (4),
either 6 or 5 would have to be at the apex. In either case, 8 would be in
row 3--no (5)--or row 4--also no (6). So, the second
combination of 1 and 7 across row 2 cannot work. If the numbers in row 2
were 3 and 9, since the apex and the rightmost numbers in rows 2, 3, and 4
sum to 26 (2), the numbers in the apex and rightmost in rows 3 and 4 would
be 8-7-2, 8-5-4, or 7-6-4 in some order. In the 8-7-2 case, since the 2
isn't in row 3 (1) and the 8 can't be in row 3 (5), the rightmost number
in row 3 would be 7. The 8 would be at the apex (6) and the 2 rightmost in
row 4. However, with 0 and 1 then also in row 3 (5) and one of 4, 5, and 6
leftmost in row 4, clue 4, which says that the leftmost number in row 2 is
larger than the leftmost number in row 4, could not work. In the 8-5-4
case, since the 8 isn't in row 3 (5) and the 8 isn't in row 4 (6), the
8 would be at the apex of the pyramid. The 4 couldn't be the rightmost
number in row 3, since the other two numbers in row 2 would then be 1 and 3
and the 3 would be in row 2; 5 would be rightmost in row 3 with 4 rightmost
in row 4. 1 and 2 would complete row 3 in some order; in either case, by
clue 4, the 0 would have to be leftmost in row 4. However, the difference
between the remaining two row 4 numbers, 6 and 7, contradicts clue 3. In the
7-6-4 case, the 4 couldn't be the rightmost number in row 3, since the other
two numbers in row 2 would then be 1 and 3 and the 3 would be in row 2.
If 7 were rightmost in row 3 with 0 and 1
the other two numbers in row 3 (5), by clue 4 the leftmost number in row 4
would be 2, leaving 5 and 8 as the middle two numbers in row 4 and making
clue 3 impossible to satisfy. If 6 were rightmost in row 3 with 0 and 2
the other two numbers in row 3 (5), by clue 4 the leftmost number in row 4
would be 1, again leaving 5 and 8 as the middle two numbers in row 4 and
making clue 3 impossible to satisfy. So, the combination of 3 and 9 in
row 2 cannot work--row 2 has 2-8 across. Since the apex and the rightmost
numbers in rows 2, 3, and 4 sum to 26 (2), the numbers in the apex and
rightmost in rows 3 and 4 must be be 9-6-3, 9-5-4, or 7-6-5 in some order.
In the 9-5-4 case, the 9 couldn't be the rightmost number in row 3 (5).
If 5 were the rightmost number in row 3, the other two numbers in the
row would be 0 and 3 in some order. By clue 4, then, 1 would be leftmost
in row 4, leaving 6 and 7 as the middle two numbers in row 4 and
contradicting clue 3. If 4 were the rightmost number in row 3, the other
two numbers in the row would be 1 and 3 in some order. By clue 4, then,
0 would be leftmost in row 4, again leaving 6 and 7 as the middle two numbers
in row 4 and contradicting clue 3. In the 7-6-5 case, 9 couldn't be in row
3 (5) or leftmost in row 4 (4). 9 would have to be the 3rd number across
row 4, but 7 would have to be the second number (3)--no. So, the three
numbers in the apex and rightmost in rows 3 and 4 are 9-6-3 in some order.
The 9 isn't at the apex (4) or rightmost in row 3 (5); it is rightmost in
row 4. Since the three numbers in row 3 sum to 8 (5), 6 can't be rightmost
in row 3 because the other two numbers would have to be 0 and 2. 6 is at
the apex and 3 is rightmost in row 3. Either 0-5 or 1-4 are the other
numbers in row 3. If 0 and 5 were the other row 3 numbers, 1 would be
leftmost in row 4 (4), leaving 4 and 7 as the middle two numbers in the
bottom row--impossible (3). 1 and 4 are the other two numbers in row 3,
with the 1 leftmost and the 4 in the middle (7). By clue 4, the leftmost
number in row 4 is 0. By clue 3, 5 is second and 7 is third across row 4.
In sum, Number Pyramid 12 is filled as follows:
|
|
|
