Richard took glass, metal and plastic items to the recycling plant and received 1 "green point" for every 4 items in each category that he brought in. In any category, items in excess of a multiple of four were not counted, so, for example, if he brought in 21 glass items, 30 metal items and 43 plastic items, he'd earn [21/4] + [30/4] + [43/4] = 5 + 7 + 10 = 22 green points. (The [] square brackets indicate the floor function--the greatest integer not exceeding the value within.)

One week he brought in a 2-digit number of glass items, a larger 2-digit number of metal items and a still larger 2-digit number of plastic items. For each class of item he received a 1-digit number of green points.

Among the three 2-digit numbers and three 1-digit numbers involved, all the non-zero digits, 1 through 9, appeared exactly once.

If I told you the total number of items brought, you'd be able to deduce how many were glass, how many were metal and how many were plastic.

How many of each category were there?

I used the following code

CLS 0
FOR n1d1 = 1 TO 9
 FOR n1d2 = 1 TO 9
  IF n1d2 <> n1d1 THEN
   glass = 10 * n1d1 + n1d2
   gp1 = INT(glass / 4)
   IF gp1 < 10 AND gp1 > 0 AND gp1 <> n1d1 AND gp1 <> n1d2 THEN
    FOR n2d1 = 1 TO 9
     IF n2d1 <> n1d1 AND n2d1 <> n1d2 AND n2d1 <> gp1 THEN
      FOR n2d2 = 1 TO 9
       IF n2d2 <> n1d1 AND n2d2 <> n1d2 AND n2d2 <> n2d1 AND n2d2 <> gp1 THEN
        metal = 10 * n2d1 + n2d2
        gp2 = INT(metal / 4)
        IF metal > glass AND gp2 > 0 AND gp2 < 10 AND gp2 <> n1d1 AND gp2 <> n1d2 AND gp2 <> n2d1 AND gp2 <> n2d2 AND gp2 <> gp1 THEN
         FOR n3d1 = 1 TO 9
          IF n3d1 <> n1d1 AND n3d1 <> n1d2 AND n3d1 <> n2d1 AND n3d1 <> n2d2 AND n3d1 <> gp1 AND n3d1 <> gp2 THEN
           FOR n3d2 = 1 TO 9
            IF n3d2 <> n1d1 AND n3d2 <> n1d2 AND n3d2 <> n2d1 AND n3d2 <> n2d2 AND n3d2 <> n3d1 AND n3d2 <> gp1 AND n3d2 <> gp2 THEN
             plastic = 10 * n3d1 + n3d2
             gp3 = INT(plastic / 4)
             IF plastic > metal AND gp3 > 0 AND gp3 < 10 AND gp3 <> n1d1 AND gp3 <> n1d2 AND gp3 <> n2d1 AND gp3 <> n2d2 AND gp3 <> n3d1 AND gp3 <> n3d2 AND gp3 <> gp1 AND gp3 <> gp2 THEN
              tot = glass + metal + plastic
              PRINT glass, metal, plastic, tot
             END IF
            END IF
           NEXT n3d2
          END IF
         NEXT n3d1
        END IF
       END IF
      NEXT n2d2
     END IF
    NEXT n2d1
   END IF
  END IF
 NEXT n1d2
NEXT n1d1

and got these four results

glass        metal        plastic       total

 16            29            35            80
 17            25            38            80
 18            25            37            80
 19            27            35            81

and of these only 19,27,35 results in a unique total

thus there were 19 glass 27 metal and 35 plastic items for a total of 81

Posted by Daniel on 2009-05-31 13:07:45