There are six boxes containing 5, 7, 14, 16, 18, 29 balls of either red or blue in colour. Some boxes contain only red balls and others contain only blue.

One sales man sold one box out of them and then he says, "I have the same number of red and blue balls left over."

Which box is sold out?

After you remove the one box, the remaining total must be able to be split evenly in half; therefore there must be an even total number of balls left. The overall total number is 89, so the box removed must have an odd number of balls in it, eliminating half of the possibilites.

The first thing I then tried in my head, removing 29 to leave 60, showed that 14+16 and 5+7+18 indeed both add up to 30. Further inspection shows that it doesn't work out if the box with either 5 or 7 balls is removed.

Other than knowing that the box has an odd number of balls in it, I'm not sure how this would be approached other than simple trial and error (not that it's that hard).

Posted by DJ on 2003-08-15 07:34:02