There are nine shiny new lottery balls in a lottery machine. They are numbered 1, 2, 3, 4, 5, 6, 7, 8, and 9. The machine dispenses one ball to each of the three people. **Apple**, **Bean**, and **Cherry**. Each person knows only the number of their own ball. They do not know the numbers of the balls that the others were given. Nor do they know which numbers were left in the machine.

Before the game begins, each of them shows their lottery ball to a fourth person, Zog. Zog then says, “One of the lottery balls is the sum of the numbers on the other two balls.”

At which point the following conversation ensues….

Apple: “There are 8 possibilities for Bean’s lottery ball.”

Bean: “There are 8 possibilities for Cherry’s lottery ball.”

Cherry: “There are 4 possibilities for Apple’s lottery ball.”

Apple: “I know Bean’s lottery ball!”

Bean: “I know Cherry’s lottery ball!”

*Question: What is Cherry’s lottery ball?*

*Question: What is Cherry’s lottery ball?*

Did you manage to solve it? For those that didn’t (and even for those that did) here is the solution…

**Solution**

The first challenge here is to figure out why Zog’s statement is relevant.

If the sum of two of the balls is equal to the third, then we can deduce that it is impossible for *both* the 8 and the 4 to be among the three selected balls. This is because if the 8 and the 4 were selected, the third ball would have to also be 4. So, 4 + 4 = 8, but this is impossible since there is only one 4-ball.

So, if Apple had either the 8, she would know that Bean does not have the 4. If she had the 4, she would know Bean does not have the 8. In either case, she would deduce that there are only seven possibilities for Bean’s lottery ball. It could be any ball with the exception of the 8-ball and the 4-ball.

In the question, Apple says there are 8 possibilities for Bean’s lottery ball. We know she does not have either the 4 or the 8.

For the same reasons, when Bean says there are 8 possibilities for Cherry’s lottery ball, we can deduce he doesn’t have the 4 or the 8 either.

Now to Cherry. She knows that her two pals don’t have the 4 or the 8. If she has the 4, or the 8, she would state that there are seven possibilities for Apple’s lottery ball. All the balls except the 4-ball and the 8-ball. But she states there are four possibilities, so we can rule out the chance she has the 4 or the 8.

So, no one has the 4 or the 8, which means the three lottery players must have balls from the following selection: 1, 2, 3, 5, 6, 7, and 9.

From Zog’s statement we can now deduce that these combinations are also impossible: (1,9), (2,6), (5, 9), and (3, 7). In each of these cases, it is impossible for there to be a third ball such that the number on a ball is equal to the sum of the numbers on the others.

Therefore, if Cherry has a 1, there are five possibilities for Apple (all those except 1, 4, 8, and 9). If Cherry has a 2, there are also five possibilities for Apple (all those except 2, 4,8, and 6), and so on. It is clear that only if Cherry has the 9-ball, will she know that there are 4 possibilities that Apple can have. These are 2, 3, 6, and 7, (i.e all those except 1, 4, 5, 8, and 9).

The question asked us to find Cherry’s lottery ball – which is 9!

Apple knows Bean’s ball since its number is 9 minus the number on her ball, but neither Cherry nor us puzzlers can deduce Bean’s ball. That then is the mystery of Cherry’s lottery ball. Solved!

*This puzzle was taken from a post written by **Alex Bellos** **@alexbellos** which first appeared in **the Guardian newspaper** on Mon 13 Aug 2018 07.10 BST. Last modified on Mon 13 Aug 2018 18.52 BST.*

*If you are interested in problem-solving and would like to find out if you are smarter than a Singaporean 10-year-old, then you may be interested in** **Can You Solve My Problems?** : A casebook of ingenious, perplexing, and totally satisfying puzzles** by Alex Bellos and available at** the Book Depository**.*