Five pumpkins are weighed two at a time in all possible combinations, similar to the first pumpkins puzzle. The results of the weighings gives nine different values: 52, 56, 60, 68, 72, 76, 80, 84, and 88. One of the values was repeated, but which value was not written down.
Find out the weights of the pumpkins and which value is the repeat.

DATA 52, 56, 60, 68, 72, 76, 80, 84, 88

CLS

FOR i = 1 TO 9
 READ w(i)
 t = t + w(i)
NEXT

max = (t + w(9)) / 20
min = (t + w(1)) / 20

PRINT min, max

FOR w1 = 0 TO 26
  lg = 80 - w1 ' arbitrary guess
  w2 = 52 - w1
  FOR w3 = w2 + 2 TO lg STEP 2
  FOR w4 = w3 + 2 TO lg STEP 2
  FOR w5 = w4 + 2 TO lg STEP 2
   IF w4 + w5 = 88 THEN
     REDIM ttl(111)
     REDIM ttl2(111)
     ttl(w1 + w3) = 1
     ttl(w1 + w4) = 1
     ttl(w1 + w5) = 1
     ttl(w2 + w3) = 1
     ttl(w2 + w4) = 1
     ttl(w2 + w5) = 1
     ttl(w3 + w4) = 1
     ttl(w3 + w5) = 1
     ttl2(w1 + w3) = ttl2(w1 + w3) + 1
     ttl2(w1 + w4) = ttl2(w1 + w4) + 1
     ttl2(w1 + w5) = ttl2(w1 + w5) + 1
     ttl2(w2 + w3) = ttl2(w2 + w3) + 1
     ttl2(w2 + w4) = ttl2(w2 + w4) + 1
     ttl2(w2 + w5) = ttl2(w2 + w5) + 1
     ttl2(w3 + w4) = ttl2(w3 + w4) + 1
     ttl2(w3 + w5) = ttl2(w3 + w5) + 1
     IF ttl(56) AND ttl(60) AND ttl(68) AND ttl(72) AND ttl(76) AND ttl(80) AND ttl(84) THEN
      PRINT w1; w2; w3; w4; w5
      FOR i = 1 TO 111
        IF ttl2(i) = 2 THEN PRINT i
      NEXT
     END IF
   END IF
  NEXT
  NEXT
  NEXT
NEXT

produces

 34.4          36.2
 20  32  36  40  48
 68

the first two numbers being merely the lowest possible average and highest possible average for the pumpkins' weights, used in initial thinking about the problem.

Posted by Charlie on 2005-09-08 20:08:23