Absolute values of 1-10 in a pyramid form

Question

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With the numbers 1-10, use each number once so that any number is equal to the absolute value of the difference of the upper two numbers. any ideas?

Answer 1

I am assuming the question is to write four of the numbers from 1 to 10 in a row, then in the three gaps write the absolute value of the difference of the two numbers above, then two, then one. Each number should appear once in the final diagram. So ignoring the restriction to 1 through 10 a possible pattern would be

12    8     3     19
   4      5    16
      1     11
        10

I have it-a couple thoughts: 10 must be in the top row, as must 9 unless 1 is next to 10. There aren't that many possibilities: 4 (places for 10) * 9 * 8 * 7 /2 (for mirror image)=1008 choices.

Answer 2

I suspect the problem is this. Fill in the triangle

a b c d

. e f g

. . h i

. . . j

with the numbers 1 to 10, using each once, so that each number in the last three rows is the absolute value of the difference of the two above it, i.e. e = |a - b|, f = |b - c|, ..., j = |h - i|.

There are, by my count, 8 solutions (of course the left-to-right reflection of a solution is a solution).

Answer 3

(Too long for a comment.)

A generalization: If $n$ is a triangular number (i.e., $n = \frac{m(m+1)}{2}$ for some positive integer $n$) you can ask the question of whether it is possible to place $1, 2, \ldots, n$ to form such a triangle. Such triangles are called exact difference triangles. OP's question is for $n = 10$ and $m = 4$.

It has been proved (by G. J. Chang, M. C. Hu, K. W. Lih and T. C. Shieh in "Exact Difference Triangles," Bulletin of the Institute of Mathematics, Academia Sinica, Taipei, Taiwan (vol. 5, June 1977, pages 191- 197)) that exact difference triangles are only possible for for $m \leq 5$.

See also Gerry Myerson's answer to this MO question: "Integers in a triangle, and differences."

Answer 4

I have solution that I got by some trial and error. These are the heuristic rules I used. They are crude, but not useless

1) Proceed from bottom to top.

2) At each stage, keep track of the number of odd and even numbers that have been used up. You start with 5 odd and 5 even numbers. If the lowest element is odd, there must be one odd and one even number above it. Proceed from bottom to top keeping track of this.

3) 10 must be on the top row.

4) If you encounter a level where you cannot fill in any numbers in a consistent manner, backtrack to the previous level and make changes. You might have to do quite a bit of backtracking.

The solution I got is

6  10  1  8
 \/ \/ \/
 4  9  7
  \/ \/
  5  2
   \/
   3