+---+                         +---+
          | 9 |→ 9                      |   |→ 9 
          +---+                         +---+
          | 8 |→ 8                      |   |→ 8
          +---+                         +---+   
          | 1 |→ 7                      |   |→ 7
          +---+                         +---+  
          | 0 |→ 6                      |   |→ 6
          +---+                         +---+ 
          | 5 |→ 5                      |   |→ 5
          +---+                         +---+
          | 4 |→ 4                      |   |→ 4
          +---+                         +---+  
    ↑     | 3 |→ 3                ↑     |   |→ 3
+---+---+ +---+               +---+---+ +---+
| 7 | 2 | | 6 |→ 2            |   |   | |   |→ 2
+---+---+ +---+               +---+---+ +---+

As the elevator rises along the eight-floor tower, it forms a series of three- digit numbers by combining the 72 in the elevator with the digit on the floor.

What's more, these three-digit numbers are multiples of 2, 3, 4, etc., up to 9. (That is, on the lowest floor, 726 is evenly divisible by 2, on the next floor, 723 is evenly divisible by 3, and so on.)

Find another arrangement for the digits 0 to 9 (using each digit exactly once, one digit per box) so that the elevator isn't 72 and the combinations of the elevator with the level form an appropriate multiple.

DECLARE SUB addOn (n!)
CLEAR , , 25000

DIM SHARED h(2 TO 9), used(9), el, ct

OPEN "towrtrav.txt" FOR OUTPUT AS #2

CLS
FOR el = 10 TO 99
  IF el <> 72 THEN
    used(el \ 10) = 1: used(el MOD 10) = 1
    addOn 2
    used(el \ 10) = 0: used(el MOD 10) = 0
  END IF
NEXT el

PRINT ct

used(7) = 1: used(2) = 1
el = 72: addOn 2
used(7) = 0: used(2) = 0

CLOSE
PRINT ct

SUB addOn (n)
 
  elv = 10 * el
  FOR i = 0 TO 9
    IF used(i) = 0 AND (elv + i) MOD n = 0 THEN
      h(n) = i
      used(i) = 1

      IF n = 9 THEN
        PRINT #2, el,
        FOR j = 2 TO 9
          PRINT #2, h(j);
        NEXT j: PRINT #2,
        ct = ct + 1
      ELSE
        addOn n + 1
      END IF

      used(i) = 0
    END IF
  NEXT
END SUB

finds a unique solution not using that 72:

 53            8  7  2  0  4  9  6  1

 
but the given pattern for 72 is only one of five:

 
 72            0  3  4  5  6  1  8  9 
 72            4  3  0  5  6  1  8  9 
 72            4  3  8  5  6  1  0  9 
 72            6  3  4  5  0  1  8  9 (the one in the puzzle)
 72            8  3  4  5  6  1  0  9 

Posted by Charlie on 2013-05-10 17:25:51