2

I have encountered a procedure that produces the magic constant 34 for a 4x4 grid with the numbers 1-16 arranged in consecutive order, and I can't figure out why it works. It is as follows:

  1. Arrange the numbers 1 to 16 into a 4 by 4 grid.
  2. Choose a number and circle it.
  3. Cross out the numbers which are in the same row as your chosen number and the numbers in the same column as your chosen number. (An image may make this instruction more clear.)
  4. Repeat this process twice more, choosing from the remaining numbers.
  5. Circle the last remaining number.
  6. Add the four circled numbers.

I have explored this site and the web and found much information about magic squares, but nothing that explains this. I can see that no circled number will be in the same row or the same column as another, which seems like it would have something to do with it, but I haven't been able to get beyond that.

I am a schoolteacher trying to help a smart middle-schooler work through this (and learn about magic squares in general), so an ideal explanation would not require math beyond algebra.

Wallfacer
  • 23
  • My intuition tells me that the sum of those four number is equal to the sum of the averages of each row/column. Which indeed is equal to 34. Moreover, it is four times the average of all number combined. – T C Molenaar Nov 05 '21 at 15:53
  • 1
    Is your student familiar with number bases? Converting to base 4 would make this very obvious that $11_4+22_4+33_4+44_4 = 34$. (This is essentially the algebra argument). – Calvin Lin Nov 05 '21 at 15:57

2 Answers2

3

The number in row $r$ and column $c$ is $4r + c - 4$.

You pick out four numbers, one in each row and one in each column. Say those are the numbers in row $r_i$ and column $c_i$, for i = 1, 2, 3, 4.

So their sum is $4(r_1+r_2+r_3+r_4) + (c_1+c_2+c_3+c_4) - 4 \times 4$.

Now $r_1 + r_2 + r_3 + r_4 = 1 + 2 + 3 + 4 = 10$ because the row numbers are 1 through 4 in some order; similarly for the $c_i$. So the sum is $4 \times 10 + 10 - 16 = 34$.

Michael Lugo
  • 22,354
2

Two preliminaries:

  1. Choose any rectangle on the magic square. Note that the sum of two
    opposite corners is always the same as the sum of the other two opposite corners. It should be clear why this is so.

  2. Add up the numbers on the main diagonal. The total is 34. (Note this is 4 times the average element.)

Now, your procedure picks four elements, all of which are on different rows and columns. One possible configuration is to take the main diagonal - which we already know sums to 34. But you can switch between any two such configurations by drawing rectangles and swapping one set of opposite corners for the other. So you always have 34. We could say that this total is an invariant on swapping pairs of opposite corners.

Blitzer
  • 2,018