There are six boxes containing 5, 7, 14, 16, 18, and 29 balls of either red or blue in colour.
Some boxes contain only red balls and the rest of them contain only blue.
A salesman sold one of the six boxes, and then he says, "I have the same number of red and blue balls left over."
Which box was sold?
Solution:
The box containing 29 balls was the one that was sold.
The six boxes contain a total of 29 + 18 + 16 + 14 + 7 + 5 = 89 balls, which is an odd number.
Now, after selling a particular box, the respective numbers of red and blue balls are equal.
Accordingly, it follows that the box sold must contain an odd number of balls.
Thus, the box that was sold must be either 5, 7, or 29.
Case 1:
If the box sold contained 5 balls, then that leaves us with 84 balls, 42 red and 42 blue.
However, it is not possible to group the remaining balls so that you get two sums of 42.
For example, one of the groups must contain 29 balls, and there are no other numbers which would add up to 13 (to get 42).
Case 2:
If the box sold contained 7 balls, then that leaves 82 balls, 41 red and 41 blue.
However, it is not possible to group the remaining balls so that you get two sums of 41.
For example, one of the groups must contain 29 balls, and there are no other numbers which would add up to 12 (to get 41).
Case 3:
If the box sold contains 29 balls, then that leaves us with 60 balls, 30 red and 30 blue.
This works: 5 + 7 + 18 = 14 + 16 = 30.
Consequently, the salesman must have sold the box containing 29 balls.
Correctly solved by:
| 1. Dr. Hari Kishan |
D.N. College, Meerut, Uttar Pradesh, India |
| 2. Davit Banana | Istanbul, Turkey |
| 3. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 4. Kamal Lohia |
Holy Angel School, Hisar, Haryana, India |
| 5. Ivy Joseph | Pune, Maharashtra, India |
| 6. Ryan Hall |
Parkview Elementary School, Chico, California, USA |
| 7. Rod Fletcher | Tainan, Taiwan |