Albert and Bernard became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

```May 15      May 16      May 19
June 17     June 18
July 14     July 16
Aug 14      Aug 15      Aug 17
```

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.
Bernard: At first I don’t know when Cheryl’s birthday is, but I know now.
Albert: Then I also know when Cheryl’s birthday is.

When is Cheryl’s birthday?

Answer

Cheryl’s birthday is on July 16.

From Albert’s first statement, we can eliminate May and June. The dates May 19 and June 18 are unique (19 only appears in May 19 and 18 only appears in June 18) and if there is no way for Bernard to know Cheryl’s birthday, then it cannot be May or June.

Bernard can now eliminate May and June after Albert’s first statement. If Bernard can now deduce Cheryl’s birthday, the date has to be unique. Therefore, we can eliminate July 14 and Aug 14 as well. Now we are left with July 16, Aug 15 or Aug 17.

Albert now knows it can only be July 16, Aug 15 or Aug 17. He knows the month but not the date, and that is enough to tell him the correct answer. Therefore it can only be July 16. If it had been August, then he will not know if it is Aug 15 or Aug 17.

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