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Number Pyramid 12
This 5-star toughie adds up with much hard work.
May 19, 2008
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:

        6
       2 8
      1 4 3
     0 5 7 9

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