Skip to content

Cheryl’s birthday

April 22, 2015

In the last week or so, there has been quite a buzz in some circles about a logic problem given to 14 year old students for the Singapore and Asian Schools Math Olympiad (SASMO) for Secondary 3 and 4 students, as follows:

Albert and Bernard just 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      
August 14 August 15   August 17    

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively, to which Albert and Bernard respond by saying:

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

So when is Cheryl’s birthday?

Answer

The key is to be clear at all times about what Albert and Bernard each know, and what we know as a result of hearing what they say.

Initially:
– Albert knows that the month of Cheryl’s birthday is either May, June, July or August.
– Bernard knows that the day of Cheryl’s birthday is either 14, 15, 16, 17, 18 or 19.
– each knows what the other knows.

  1. Albert: “I don’t know when Cheryl’s birthday is”.
    Given Cheryl’s ten supplied dates, every one of the four months has multiple possible days. So, before Albert told us, both we and Bernard already knew this.
  2. Albert: “I know that Bernard does not know too.”
    To say this, Albert must know that May 19 and June 18 are not possible dates for Cheryl’s birthday, otherwise Bernard would know the date of Cheryl’s birthday immediately. So, given that Albert knows only the month of Cheryl’s birthday, the month cannot be May or June.

So now our table of possible dates looks like this…

  May 15 May 16      May 19 
      June 17 June 18   
July 14   July 16      
August 14 August 15   August 17    
  1. Bernard: “At first I didn’t know when Cheryl’s birthday is.”
    Albert already knew this, and has just told us so.
  2. Bernard: “Now I do know [when Cheryl’s birthday is].”
    This means the day of Cheryl’s birthday cannot be 14, otherwise July 14 and August 14 would still be possible dates for Bernard, so he would still not know when Cheryl’s birthday is.

So now our table of possible dates looks like this…

  May 15 May 16     May 19
      June 17 June 18  
July 14   July 16      
August 14 August 15   August 17    
  1. Albert: “Then I also know when Cheryl’s birthday is.”
    If Cheryl’s birthday was in August, both August 15 and August 17 would still be possible dates for Albert, so he would still not know when Cheryl’s birthday is. So Cheryl’s birthday is not in August.

So now our table of possible dates looks like this…

  May 15 May 16     May 19
      June 17 June 18  
July 14th   July 16      
August 14  August 15   August 17    

So Cheryl’s birthday is on July 16.

Advertisements
Leave a Comment

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: