A "Number Pyramid" is composed of ten different numbers ( usually 0  9 ) with four rows. ONE EXAMPLE of a Number Pyramid is as follows:
0
1 2
3 4 5
6 7 8 9
Given the clues below, determine the composition of a new Number Pyramid.
1) The sum of the two numbers in the second row is 11.
2) The sum of the numbers in the bottom row minus the sum of the numbers in the third row equals 10.
3) The rightmost numbers in the four rows must sum to 18.
4) The middle number in the third row minus the leftmost number in the second row should equal 4.
5) Subtracting the top number in the Pyramid from the rightmost number in the second row leaves 4.
6) In the bottom row, the leftmost number is greater than the second number from the left, while the rightmost number is greater than the second number from the right.
For bonus marks, which of the above clues (if any), are not necessary in order to construct the pyramid?
A
B C
D E F
G H I J
Initial assumption: we are dealing with numbers 09.
Clue 1 tells us that the second row is either 5&6, 4&7, 3&8 or 2&9. So B = 2 through 9, and C = 11B.
Clue 4 tells us that EB=4. This means that E must be 4 or greater, and B must be 5 or less. Combine this with the information from Clue 1 and we see B can only be 2 through 5, so E can only be 6 through 9. Also, C can only be 6 through 9.
Clue 5 is very similar to Clue 4. In this case, it tells us that CA = 4. This means that C must be 4 or greater (which we already knew), and A must be 5 or less. Combine this with the information from Clue 1 and we see again that C can only be 6 through 9, so A must be 2 through 5.
Also, EB=4 means B=E4. And CA=4 means C = A+4. So C = 11B means A+4 = 11(E4) = 11E+4 which means A = 11E or A+E = 11.
Let's think about Clue 2. So (G+H+I+J)  (D+E+F) = 10. Well, the most that the fourth row could be is 30 (9+8+7+6). But in that case, the most the third row could be is 12 (5+4+3). This is a difference of 18 which is too big. Moving on in this fashion, we'll see that the fourth row can be at most 26 (in which case the third row is at most 16). If the fourth row were greater than 26, the third row would not be able to be 10 less than it. So we have an upper bound. Let's think about a lower bound.
At first glance, the least that the third row could be is 3 (0+1+2). But remember that E must be 6, 7, 8, or 9. So really, the least that the third row can be is 7 (0+1+6). In this case, the least that the fourth row can be is 14. This is a difference of 7. So there aren't any number restrictions that would force the lower bound of the third row to be greater than 7. But this automatically forces the lower bound of the fourth row to be 17.
Well, considering the fact that with no restrictions, the fourth row can be 630 and the third row could be 324, I think this is a pretty good start for narrowing things down.
Moving on.
Let's look at Clue 3. What are all the ways of getting 18 with four numbers? Maybe once we look at them we'll be able to find some that can be eliminated. If the sum of any two numbers is 11 we may have a problem. If A or C have no choice but to be one of the addends for the 11 (based on their restrictions), then E or B (which are not one of these 4 numbers summing to 18) would have to be the other addend for the 11, which causes a problem. Also, if A or C can only be ONE of the numbers (based on their restrictions), the corresponding C or A that satisfies CA=4 must be present. Otherwise it is not a solution.
Based on this, I was able to get rid of 12 of the 17 possible cases. I hope I didn't miss any!
Case 1: 0+4+5+9 : C must be 9 (B must be 2), A must be 5 (E must be 6), 0 must be F (I can't be <J=0 for Clue 6), J must be 4.
Case 2: 0+4+6+8 : A must be 4 (E must be 7), C must be 8 (B must be 3), 0 must be F (I can't be <J=0 for Clue 6), J must be 6.
Case 3: 1+3+5+9 : C must be 9 (B must be 2), A must be 5 (E must be 6), 1 must be F (see note), J must be 3.
Case 4: 1+4+5+8 : C must be 8 (B must be 3), A must be 4 (E must be 7).
Case 5: 2+3+6+7 :
Note: In Case 3, if the 1 is in the fourth row (J=1) then I=0 (for Clue 6). The fourth row can only be 1726. So let's see if there are any ways for 0 and 1 to be used to get 1726. The largest number that can be made with 0 and 1 is 18.
0+1+8+9 = 18, but 9 is already taken.
0+1+7+9 = 17, but 9 is already taken.
So the J can't equal 1 in that case.
Cases 1 and 3 require C = 9. Let's see where that takes us. If we find a contradiction, we might be able to knock off two cases at once. So, C=9, B=2, E=6, A=5.
In Case 1:
F=0 and J=4, and the numbers left are 1, 3, 7, 8.
If D = 1, then D+E+F = 7, and G+H+I+J = 3+7+8+4 = 22. Too big.
If D = 3, then D+E+F = 9, and G+H+I+J = 1+7+8+4 = 20. Too big.
If D = 7, then D+E+F = 13, and G+H+I+J = 1+3+8+4 = 16. Too small. Making D = 8 will just make G+H+I+J smaller. So Case 1 does not work.
In Case 3:
F=1 and J=3, and the numbers left are 0, 4, 7, 8.
Then I=0 since nothing else is less than 3.
If D=4, then D+E+F = 11, and G+H+I+J = 7+8+0+3 = 18. Too small. Making D = 7 or 8 will just make G+H+I+J smaller. So Case 3 does not work.
Next let's look Cases 2 and 4. So C=8, B=3, E=7, A=4.
In Case 2:
F=0 and J=6, and the numbers left are 1, 2, 5, 9.
If D = 1, then D+E+F = 8, and G+H+I+J = 2+5+9+6 = 22. Too big.
If D = 2, then D+E+F = 9, and G+H+I+J = 1+5+9+6 = 21. Too big.
If D = 5, then D+E+F = 12, and G+H+I+J = 1+2+9+6 = 18. Too small. Making D = 9 will just make G+H+I+J smaller. So Case 2 does not work.
In Case 4:
F and J are either 1 or 5, and the number left are 0, 2, 6, 9.
If J=1, then I=0 and F=5. Then if D = 2, D+E+F = 14 and G+H+I+J = 6+9+0+1 = 16. Too small. Making D bigger will just make G+H+I+J smaller. So Case 4 with J=1 doesn't work.
Let's look at J=5 and F=1. If D = 0, then D+E+F = 8, and G+H+I+J = 2+6+9+5 = 22. Too big.
If D = 2, then D+E+F = 10, and G+H+I+J = 0+6+9+5 = 20. This works!
So one solution is:
4
3 8
2 7 1
9 6 0 5
The only case I haven't checked yet is Case 5. For completeness...
We either have C=7, A=3, B=4, and E=8 (with 0, 1, 5, 9 left), or we have C=6, A=2, B=5, and E=9 (with 0, 1, 4, 8 left).
C=7, A=3, B=4, and E=8 (with 0, 1, 5, 9 left).
If J=2 and F=6:
If D = 0, then D+E+F = 14, and G+H+I+J = 1+5+9+2 = 21. Too small. Making D bigger will just make G+H+I+J smaller. So this case does not work.
If J=6 and F=2:
If D = 0, then D+E+F = 10, and G+H+I+J = 1+5+9+6 = 21. Too big.
If D = 1, then D+E+F = 11, and G+H+I+J = 0+5+9+6 = 20. Too small. Making D bigger will just make G+H+I+J smaller. So C=7 does not work.
C=6, A=2, B=5, and E=9 (with 0, 1, 4, 8 left).
If J=3 and F=7:
If D = 0, then D+E+F = 16, and G+H+I+J = 1+4+8+3 = 16. Too small. Making D bigger will just make G+H+I+J smaller. So this case does not work.
If J=7 and F=3:
If D = 0, then D+E+F = 12, and G+H+I+J = 1+4+8+7 = 20. Too small. Making D bigger will just make G+H+I+J smaller. So C=6 does not work.
So Case 5 doesn't work at all, and the one solution I found is the only one. And I didn't find anything that contradicted my initial assumption.
(fixing formatting typos)
Oh yeah, I needed all the clues, and I didn't see anything that was redundant.
Edited on August 20, 2004, 11:37 am

Posted by nikki
on 20040820 11:24:28 