1. How many primes are in a clock if you regard the numbers on its face as a continuous clockwise string of digits; not to exceed 15 digits i.e. one full round?
2.Same question for a counterclockwise direction.
3.Same question for a digital watch ((HH(0-23)MM(0-59)) - in ascending order).
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DECLARE SUB factor (num#, s$)
DEFDBL A-Z
src$ = "123456789101112123456789101112"
FOR l = 1 TO 15
FOR st = 1 TO 15
s$ = MID$(src$, st, l)
n = VAL(s$)
factor n, s1$
IF LTRIM$(s1$) = s$ THEN ct = ct + 1: PRINT ct, n
NEXT
NEXT l
ct = 0
src$ = "211101987654321211101987654321"
FOR l = 1 TO 15
FOR st = 1 TO 15
s$ = MID$(src$, st, l)
n = VAL(s$)
factor n, s1$
IF LTRIM$(s1$) = s$ THEN ct = ct + 1: PRINT ct, n
NEXT
NEXT l
SUB factor (num, s$)
s$ = "": n = ABS(num): IF n > 0 THEN limit = SQR(n): ELSE limit = 0
IF limit <> INT(limit) THEN limit = INT(limit + 1)
dv = 2: GOSUB DivideIt
dv = 3: GOSUB DivideIt
dv = 5: GOSUB DivideIt
dv = 7
DO UNTIL dv > limit
GOSUB DivideIt: dv = dv + 4 '11
GOSUB DivideIt: dv = dv + 2 '13
GOSUB DivideIt: dv = dv + 4 '17
GOSUB DivideIt: dv = dv + 2 '19
GOSUB DivideIt: dv = dv + 4 '23
GOSUB DivideIt: dv = dv + 6 '29
GOSUB DivideIt: dv = dv + 2 '31
GOSUB DivideIt: dv = dv + 6 '37
IF INKEY$ = CHR$(27) THEN s$ = CHR$(27): EXIT SUB
LOOP
IF n > 1 THEN s$ = s$ + STR$(n)
EXIT SUB
DivideIt:
DO
q = INT(n / dv)
IF q * dv = n AND n > 0 THEN
n = q: s$ = s$ + STR$(dv): IF n > 0 THEN limit = SQR(n): ELSE limit = 0
IF limit <> INT(limit) THEN limit = INT(limit + 1)
ELSE
EXIT DO
END IF
LOOP
RETURN
END SUB
first finds the primes in the forward direction of the digits:
1 2
2 3
3 5
4 7
5 2
6 23
7 67
8 89
9 11
10 11
11 101
12 4567
13 67891
14 89101
15 10111
16 789101
17 4567891
18 23456789
19 56789101
20 1234567891
21 45678910111
22 12345678910111
While the above counts 22 primes, the prime 2 appears twice (once from the solitary 2 on the clock face and once from the 12), and the prime 11 appears twice (once from 11 itself and once from the end of 11 and the beginning of 12), so only 20 distinct primes actually appear.
The same holds in the reverse clockface:
1 2
2 7
3 5
4 3
5 2
6 11
7 11
8 19
9 43
10 211
11 101
12 2111
13 1019
14 1987
15 76543
16 21211
17 101987
18 432121
19 4321211
20 12111019
21 6543212111
22 9876543212111
Two and eleven appear here also twice each, so the 22 occurrences are just 20 distinct primes.
The forward direction had no examples of wrapping around the top of the clockface, but the reverse does, with 32121 appearing in four of the primes and 2121 in another.
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Posted by Charlie
on 2010-10-08 15:34:46 |