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3 Primes yield 12 (Posted on 2010-10-07) Difficulty: 3 of 5

No Solution Yet Submitted by brianjn    
Rating: 4.0000 (1 votes)

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Solution re(2): computer exploration -- oh, I see a mistake in my reasoning -- solution (spoiler) | Comment 3 of 6 |
(In reply to re: computer exploration -- oh, I see a mistake in my reasoning by Charlie)

I also mistakenly disallowed the digit zero from the addends.

Now I find ten solutions (rather than brian's stated 9):

pie  end  dry              total
241  109  967               1317
241  157  739               1137
263  347  701               1311
269  941  107               1317
281  107  743               1131
421  157  739               1317
521  107  743               1371
641  173  359               1173
683  347  701               1731
821  167  743               1731
 

The revised program:

DECLARE FUNCTION nodup! (x$)
DATA         101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179
DATA         181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269
DATA         271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367
DATA         373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461
DATA         463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571
DATA         577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661
DATA         673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773
DATA         787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883
DATA         887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997


DATA     11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71
DATA     73 , 79 , 83 , 89 , 97

DIM prime(143), prime2(21)

FOR i = 1 TO 143: READ prime(i): PRINT prime(i); : NEXT

PRINT

FOR i = 1 TO 21: READ prime2(i): PRINT prime2(i); : NEXT

PRINT : PRINT

FOR p1 = 1 TO 141
 pie = prime(p1): pie$ = LTRIM$(STR$(pie))
 chk1$ = LEFT$(pie$, 2)
 IF nodup(pie$) THEN
FOR p2 = 1 TO 142
 end0 = prime(p2): end0$ = LTRIM$(STR$(end0))
 IF RIGHT$(pie$, 1) = LEFT$(end0$, 1) AND nodup(chk1$ + end0$) THEN
   chk2$ = chk1$ + LEFT$(end0$, 2)
FOR p3 = 1 TO 143
   dry = prime(p3): dry$ = LTRIM$(STR$(dry))
 IF RIGHT$(end0$, 1) = LEFT$(dry$, 1) AND nodup(chk2$ + dry$) THEN
   chk3$ = chk2$ + dry$
   IF nodup(chk3$) THEN
     tot = pie + end0 + dry
     IF tot > 999 AND tot < 10000 THEN
      t$ = LTRIM$(STR$(tot))
      good = 1
      FOR i = 1 TO 4
        FOR j = 1 TO 4
          IF i <> j THEN
            ptest = VAL(MID$(t$, i, 1) + MID$(t$, j, 1))
            foundp = 0
            FOR k = 1 TO 21
              IF ptest = prime2(k) THEN foundp = 1: EXIT FOR
            NEXT
            IF foundp = 0 THEN good = 0: EXIT FOR
          END IF
        NEXT j
        IF good = 0 THEN EXIT FOR
      NEXT i

      ' check tot composite:
      comp = 0
      IF tot MOD 3 = 0 OR tot MOD 5 = 0 OR tot MOD 7 = 0 THEN comp = 1
      FOR i = 1 TO 21
        IF tot MOD prime2(i) = 0 THEN comp = 1
      NEXT
      IF comp = 0 THEN good = 0


      IF good THEN
          PRINT pie; end0; dry, tot
          ct = ct + 1: IF ct MOD 40 = 0 THEN DO: LOOP UNTIL INKEY$ > "": PRINT
      END IF
     END IF
   END IF
 END IF
NEXT p3
 END IF
NEXT p2
 END IF
NEXT p1
PRINT ct

FUNCTION nodup (x$)
  nd = -1
  FOR i = 1 TO LEN(x$) - 1
    IF INSTR(MID$(x$, i + 1), MID$(x$, i, 1)) THEN nd = 0: EXIT FOR
  NEXT
  nodup = nd
END FUNCTION

 

 

 

 


  Posted by Charlie on 2010-10-07 13:43:44
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