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?

Since each of the item quantities must be a 2 digit value that gives a 1 digit number of green points then  each quantity must be between 10 and 39 inclusively.  Next since 3 2-digit numbers and 3 1-digit numbers results in exactly 9 digits then there can be no digit duplication among the digits.  Thus the first digits of each quantity must be unique and since they must be between 10 and 39 then the first digits are 1,2 and 3.  So we have 1x, 2y, and 3z for our 3 quantities.  1x must be greater than 15 otherwise the green points for 1x would be either 2 or 3 and thus cause a digit duplicate.  That leaves 6,7,8,9 for x.  Similarly we know that y is one of 4,5,6,7,8,9.  And z is one of 4,5,6,7,8,9.

Now if x is 6 then that makes a green point of 4 then that eliminates all but 8,9 for y.  If y is 8 then that gives a green point of 7 but that only leaves 35 or 39 for 3z and these result in 8 and 9 respectively for green points and thus y can not be 8.  If y is 9 then that still gives 7 for green points.  Now we have either 35 or 38 and only 35 is valid.  Thus if x is 6 then y is 9 and z is 5 and this gives us 16,29,35 for our values and a total of 80.

Now if x is 7 then that also makes for a green point of 4.  Then that leaves y to be 5 and a green point of 6.  That only leaves 8,9 for z and only 8 is valid.  Thus we get 17,25, and 38 for our values and this also gives a total of 80.

Now if x is 8 then we get green point of 4.  Then y must be 5,7, or 9.  If y is 5 then z is 7 and we get values of 18,25, and 37 with still a total of 80.  If y is 7 or 9 this leads to no possible values for z.

Finally if x is 9 the we get green point of 4.  Then y must 7 and z must then be 5 so we get 19,27, and 35 for a total of 81.

Of all the possible quantities only 19,27,35 gives a unique total thus there were 19 glass items, 27 metal items, and 35 plastic items with 4+6+8=18 green points and a total of 81 items.

Edited on May 31, 2009, 1:42 pm

Posted by Daniel on 2009-05-31 13:30:35