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Prime clock (Posted on 2004-08-01) Difficulty: 3 of 5
Given a clock, rearrange six consecutive numbers on its face, so the sum of every pair of adjacent numbers is a prime.

See The Solution Submitted by Federico Kereki    
Rating: 3.7500 (8 votes)

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Solution The computer method | Comment 2 of 16 |

The following program cycles which 6 in a row are scrambled and goes through all 720 permutations of each six:

DECLARE SUB permute (a$)
CLS
FOR i = 1 TO 12
  face$ = face$ + CHR$(i)
NEXT
h$ = face$

prime(1) = 2
prime(2) = 3
prime(3) = 5
prime(4) = 7
prime(5) = 11
prime(6) = 13
prime(7) = 17
prime(8) = 19
prime(9) = 23

DO
 h2$ = face$
 DO
   good = 1
   f$ = face$ + LEFT$(face$, 1)
   FOR i = 1 TO 12
    p = ASC(MID$(f$, i, 1)) + ASC(MID$(f$, i + 1, 1))
    hit = 0
    FOR j = 2 TO 9
      IF p = prime(j) THEN hit = 1: EXIT FOR
    NEXT
    IF hit = 0 THEN good = 0: EXIT FOR
   NEXT
   IF good THEN
     FOR i = 1 TO 12
       PRINT ASC(MID$(f$, i, 1));
     NEXT
     PRINT
   END IF
   t$ = LEFT$(face$, 6)
   permute t$
   MID$(face$, 1, 6) = t$
 LOOP UNTIL face$ = h2$
 face$ = MID$(face$, 2) + LEFT$(face$, 1)
LOOP UNTIL h$ = face$

where the subroutine permute is shown elsewhere on this site.

The findings were

7  10  9  8  5  6  11  12  1  2  3  4
9  10  7  6  5  8  11  12  1  2  3  4

Shown starting at 1, that's

1 2 3 4 7 10 9 8 5 6 11 12  and
1 2 3 4 9 10 7 6 5 8 11 12

In one instance the 8, among the 6 rearranged numbers, is in its own position, and in the other the 7, also among the rearranged numbers, is in its own position.

 


  Posted by Charlie on 2004-08-01 11:29:46
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