A super number square has the following properties:

1. In each row, the rightmost number is the sum of the other three.
2. In each column, the bottom number is the sum of the other three.
3. Within each NW-SE diagonal line, the last number (bottom rightmost) is the product of the other numbers.

For example, if you have a square that looks like:

A B C D
E F G H
I J K L
M N P Q

you know that A+B+C=D, C+G+K=P, AFK=Q, EJ=P, and so on.

Construct a super number square in which the highest number in any position is 57, and the second number in the top row is a 5 (all numbers are positive integers).

Assuming that C must in fact be equal to H and I equal to N, we know

p=ej
l=5g
i=5g-k-j
n=5g-k-j
h=ej-k-g
c=ej-k-g
n=5g-k-j
f=ej-k-2g-e
f=5g-k-2j-5

from these last two, 7g=ej+2j-e+5

Thus various combinations of e and j produce what may or may not be valid (integral) g. We can then try various combinations of a and k to see if the product q is equal to d+h+l and to m+n+p where d=a+b+c and m is a+e+i.

Not wanting to do this by hand, I made a program:

CLS
sum = 2
DO
 FOR j = 0 TO sum
   e = sum - j
   g = (e * j + 2 * j - e + 5) / 7
     IF g = INT(g) AND g >= 0 THEN
      FOR k = 0 TO 50
       FOR a = 0 TO 50
         IF e * j - k - 2 * g - e > 0 AND 5 * g - k - j > 0 THEN
          IF a * (e * j - k - 2 * g - e) * k = a + 5 + 2 * (e * j - k - g) + 5 * g THEN
           IF a + e + 5 * g - k - j + 5 * g - k - j + e * j = a * (e * j - k - 2 * g - e) * k THEN
            PRINT USING "####"; a;
            PRINT USING "####"; 5;
            PRINT USING "####"; e * j - k - g;
            PRINT USING "####"; a + 5 + e * j - k - g

            PRINT USING "####"; e;
            PRINT USING "####"; e * j - k - 2 * g - e;
            PRINT USING "####"; g;
            PRINT USING "####"; e * j - k - g

            PRINT USING "####"; 5 * g - k - j;
            PRINT USING "####"; j;
            PRINT USING "####"; k;
            PRINT USING "####"; 5 * g

            PRINT USING "####"; a + e + 5 * g - k - j;
            PRINT USING "####"; 5 * g - k - j;
            PRINT USING "####"; e * j;
            PRINT USING "####"; a * (e * j - k - 2 * g - e) * k
            PRINT "-"
            DO: LOOP UNTIL INKEY$ > ""
           END IF
          END IF
         END IF
       NEXT
      NEXT

     END IF
 NEXT
 sum = sum + 1
LOOP

the first several resulting arrays are:

  21   5  11  37


   5   3   3  11


  11   3   1  15


  37  11  15  63


-


  19   5   9  33


   5   1   3   9


   9   3   3  15


  33   9  15  57


-


  11   5  15  31


   5   6   4  15


  15   4   1  20


  31  15  20  66


-


   9   5  10  24


   5   1   4  10


  10   4   6  20


  24  10  20  54

-
of which the second has the desired highest number of 57.

Posted by Charlie on 2003-08-19 15:53:55