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The wine is so-so, but... (Posted on 2012-04-06) Difficulty: 3 of 5
   t   h  eb
  ee   r  i
   s   o  k
Use simple substitution code (letters to digits) to get a 3x3 magic square.

Rem: eb=10*e+b

See The Solution Submitted by Ady TZIDON    
Rating: 3.7500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution | Comment 2 of 8 |

DECLARE SUB permute (a$)
CLS
a$ = "0123456789": h$ = a$
DO
  t = VAL(MID$(a$, 1, 1))
  h = VAL(MID$(a$, 2, 1))
  e = VAL(MID$(a$, 3, 1))
  b = VAL(MID$(a$, 4, 1))
  r = VAL(MID$(a$, 5, 1))
  i = VAL(MID$(a$, 6, 1))
  s = VAL(MID$(a$, 7, 1))
  o = VAL(MID$(a$, 8, 1))
  k = VAL(MID$(a$, 9, 1))
  eb = 10 * e + b
  ee = 11 * e
  IF t + h + eb = ee + r + i AND s + o + k = ee + r + i THEN
    IF t + ee + s = h + r + o AND h + r + o = eb + i + k THEN
      IF t + r + k = eb + r + s AND eb + r + s = t + h + eb AND t + h + eb = h + r + o THEN
       PRINT t; h; eb
       PRINT ee; r; i
       PRINT s; o; k
       PRINT
      END IF
    END IF
  END IF
  permute a$
LOOP UNTIL a$ = h$

finds three potential solutions, but in two of them the letter "e" represents zero and thus leads to leading zeros, so only the middle solution below is in keeping with the spirit of the puzzle:

 5  6  01
 00 4  8
 7  2  3
 6  5  10
 11 7  3
 4  9  8
 7  2  03
 00 4  8
 5  6  1

I've placed the leading zeros manually as the program prints only the single digits in those cases.


  Posted by Charlie on 2012-04-06 13:44:50
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