11

My son brought the following puzzle home from his 3rd grade math class.

enter image description here

Are there any strategies for solving it besides brute force?

Mark
  • 537
  • Try a $6$ by $4$ matrix. – Don Larynx Sep 20 '13 at 00:10
  • @Jossie Maybe you could expound a bit? – Benjamin Dickman Sep 20 '13 at 00:15
  • [ a_1, a_2, a_3, a_4; b_1, b_2, b_3, b_4; a_1 + c_2 + b_2 + c_4; a_4 + d_2 + b_3 + c_4; b_1 + c_2 + a_2 + e_4; b_4 + d_2 + a_3 + e_4; ] set all of the rows equal to 26 and solve for the linear system. – Don Larynx Sep 20 '13 at 00:28
  • 5
    @Jossie It would be interesting to see you solve it. How would you account for the fact that they must be distinct integers from 1 to 12? Also, there are 80 solutions (up to rotation/reflection). Can you explain why this is so? – Calvin Lin Sep 20 '13 at 00:42
  • To answer your second question, those solutions are just isomorphic graphs to the star of david. So it's really nothing significant. – Don Larynx Sep 20 '13 at 01:15
  • To answer your first question, notice there are 6 numbers (vertices) with degree 4 and 2, respectively. this leads to 24 + 12 = 36, the total number of variables (without replacing c_1 with a_1 for example). – Don Larynx Sep 20 '13 at 01:17
  • See please my answer to related question here: http://math.stackexchange.com/questions/499587/how-to-solve-this-system-of-linear-equations/499713#499713 – Oleg567 Sep 23 '13 at 05:03

2 Answers2

3

These kinds of problems are special because we're only looking for one solution, so that we can make educated guesses to stumble upon one.

In a magic square, 1 and the maximum number are usually paired. But in this puzzle, unlike a magic square, every number is included in exactly two sums. This gives us freedom we wouldn't normally have. We can pair 1 with 12 on the outside, but because of this freedom, we can put the next-most-extreme numbers together, to give us a place to start from: $$\begin{array}{ccccccc} & & & 1\\ ? & & ? & & 2 & & ?\\ & ? & & & & 11\\ ? & & ? & & ? & & 12\\ & & & ? \end{array}$$

There is one undetermined number in a sum with 11 and a sum with 12. That can probably be made as small as possible: 3. (We could have done this symmetrically with 1 and 2 needing a big number, but probably not both at the same time.)

$$\begin{array}{ccccccc} & & & 1\\ ? & & ? & & 2 & & ?\\ & ? & & & & 11\\ ? & & ? & & 3 & & 12\\ & & & ? \end{array}$$

To complete $3+12$, we need a sum of $11$, which we could only get as $6+5$ or $7+4$. To complete $3+11$, we need $5+7$ or $4+8$. Therefore, we can't pick $7+4$ because it would wipe out both possibilities to complete the diagonal (this feels like kakuro). As such, the row needs a 6 and a 5. Which should go in the bottom left corner? Probably the middling value 6 since it's in a sum with 1 and a sum with 12:

$$\begin{array}{ccccccc} & & & 1\\ ? & & ? & & 2 & & ?\\ & ? & & & & 11\\ 6 & & 5 & & 3 & & 12\\ & & & ? \end{array}$$

This leaves $4+8$ for the diagonal on the right. But putting 4 in the top right would make the top row have no options since the numbers would have to be too big: $$\begin{array}{ccccccc} & & & 1\\ ? & & ? & & 2 & & 8\\ & ? & & & & 11\\ 6 & & 5 & & 3 & & 12\\ & & & 4 \end{array}$$ The only pair for the top row is 7,9 and the only pair for the upper diagonal is 9 and 10, so 9 must go in the intersection: $$\begin{array}{ccccccc} & & & 1\\ 7 & & 9 & & 2 & & 8\\ & 10 & & & & 11\\ 6 & & 5 & & 3 & & 12\\ & & & 4 \end{array}$$

As made clear in How to solve this system of linear equations, this is just one of many other solutions.

Community
  • 1
Mark S.
  • 23,925
1

My sense is that this will always come down to a bit of trial and error, though there is ample discussion around this problem (including solutions) that can be found by googling.

One such solution can be seen here.

Benjamin Dickman
  • 14,451