All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Prime clock (Posted on 2004-08-01)
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 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  49  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  and1 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

 Search: Search body:
Forums (0)