Almost a Magic Square

Date: 12/16/96 at 22:38:59
From: Mike McCabe
Subject: 3rd grader stumps the family

Hi,
 
My 8-year-old son came home with a problem that we're having trouble 
with.  Is there something we're missing here or is this over the head 
of most 3rd graders?

You have 4 columns and 4 rows. List he following numbers so each row 
and column totals 44:

3,5,10,11,15,20,8,11,9,10,10,12,13,14,15,10

Is there any easy formula to arrive at the answer?  Try and try again 
is making me look bad.

Thank You,
Mike McCabe

Date: 12/17/96 at 11:03:08
From: Doctor Rob
Subject: Re: 3rd grader stumps the family

If you can create this arrangement, then you can get another one by
rearranging the rows, or by rearranging the columns.  Also, you can
get another one by making every row a column and vice versa:

  a b c d      a e i m
  e f g h ---> b f j n
  i j k l      c g k o
  m n o p      d h l p

Start with the biggest number, 20.  It must be in some row and column.
According to the previous paragraph, you might as well assume that it
appears in row 1, column 1.

Think of what sums can be made from 20 and the other numbers to add up
to 44:

  44 - 20 - 3 - 5 = 16 impossible
  44 - 20 - 3 - 8 = 13
  44 - 20 - 3 - 9 = 12
  44 - 20 - 3 - 10 = 11
  44 - 20 - 5 - 8 = 11
  44 - 20 - 5 - 9 = 10
  44 - 20 - 8 - 8 = 8 impossible

Now the row and column containing 20 must have, besides the 20, six
different numbers from the list.  Since there is only one 3, one 5,
one 8, and one 9, only the following four possibilities are left:

1)  44 = 20 + 3 + 8 + 13 = 20 + 5 + 9 + 10
2)  44 = 20 + 3 + 9 + 12 = 20 + 5 + 8 + 11
3)  44 = 20 + 3 + 10 + 11 = 20 + 5 + 8 + 11
4)  44 = 20 + 3 + 10 + 11 = 20 + 5 + 9 + 10

For each of these, we might as well assume that the first sum 
containing the 3 lies in the first row, and the second sum in the 
first column.  Since order of rows and columns is not important, we 
might as well put them in descending order:

1)  20 13  8  3   2)  20 12  9  3   3)  20 11 10  3   4)  20 11 10  3
    10  *  *  *       11  *  *  *       11  *  *  *       10  *  *  *
     9  *  *  *        8  *  *  *        8  *  *  *        9  *  *  *
     5  *  *  *        5  *  *  *        5  *  *  *        5  *  *  *

Numbers left:

  10,10,10,11,11,   10,10,10,10,11,    9,10,10,10,12,   8,10,10,11,12,  
  12,14,15,15       13,14,15,15        13,14,15,15      13,14,15,15

Now in each case the problem is reduced to a smaller one.  For 
example, in Case 1, we need to construct a 3-by-3 square of numbers 
made up from 10,10,10,11,11,12,14,15,15 with row sums 34,35,39 and 
column sums 31,36,41.  Once again, placing the largest numbers, 15 and 
15, first, we have only a few possibilities.

Continuing in this way, the solution can be found.  Mostly it is an
exercise in addition and subtraction, together with logic.  Once you
have found one, using the techniques in paragraph 1, there will be
1151 others (since 1152 = (4*3*2*1)*(4*3*2*1)*2, accounting for the
order of the rows, the order of the columns, and the swapping of rows
and columns).

It can be done!

I hope this helps.  If you are still having problems, write back.

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/