Cheryl And Betty Celebrate

Logic Level 3

One day, Cheryl and Betty list down possibilities for their birthday dates.

By some strange coincidence, they wrote the same possibilities for their birthday dates.

They were

Jun 4 Jun 6 Jun 8
Jul 4 Jul 5 Jul 6 Jul 8
Aug 4 Aug 5 Aug 7 Aug 9
Sep 3 Sep 5 Sep 7 Sep 10

Cheryl then told Betty the day of month of her birthday.

Betty told Cheryl the month of her birthday.

Cheryl: "I know that Betty does not know my birthday."

Betty: "Cheryl does not also know my birthday and my birthday is not on Sep 5."

Cheryl: "You gave me redundant information and my birthday is not on Aug 7."

Betty: "Now I know your birthday and the day of the month of my birthday is a prime number."

Cheryl: "Now I know your birthday."

What is the difference (in days) between Betty's and Cheryl's birthday? (both dates inclusive).


The answer is 65.

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2 solutions

Pi Han Goh
Mar 1, 2016

Let us first construct the tables for possibilities for their birthday dates

Cheryl’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 Betty’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 \underbrace{\text{Cheryl's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \text{August 9} \\ \text{September 3} & \text{September 5} & \text{September 7} & \text{September 10} \\ \hline \end{array}} \qquad\underbrace{\text{Betty's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \text{August 9} \\ \text{September 3} & \text{September 5} & \text{September 7} & \text{September 10} \\ \hline \end{array}}

We are told that Cheryl told Betty the day of the month of her birthday, and Betty told Cheryl the month of her birthday.

Cheryl then started off by claiming that Betty does not know her birthday. This tells us that day of the month written in the table for Cheryl's birthday isn't unique and thus has been written at least twice. This tells us that we can cancel out the dates whose day of the month only appears once (from Cheryl's possible dates):

Cheryl’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 Betty’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 \underbrace{\text{Cheryl's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \xcancel{\text{August 9}} \\ \xcancel{\text{September 3}} & \text{September 5} & \text{September 7} & \xcancel{\text{September 10} } \\ \hline \end{array}} \qquad\underbrace{\text{Betty's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \text{August 9} \\ \text{September 3} & \text{September 5} & \text{September 7} & \text{September 10} \\ \hline \end{array}}

However, with Betty's reply, we can only cancel out 5 th 5^\text{th} of September from the list of Betty's possible birthday dates as illustrated below:

Cheryl’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 Betty’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 \underbrace{\text{Cheryl's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \xcancel{\text{August 9}} \\ \xcancel{\text{September 3}} & \text{September 5} & \text{September 7} & \xcancel{\text{September 10} } \\ \hline \end{array}} \qquad\underbrace{\text{Betty's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \text{August 9} \\ \text{September 3} & \xcancel{\text{September 5}} & \text{September 7} & \text{September 10} \\ \hline \end{array}}

Afterwards, Cheryl still claims that she doesn't know Betty's birthday but she pointed out that her Birthday isn't on 7 th 7^\text{th} of August, updating Chery'ls possible bithday dates gives:

Cheryl’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 Betty’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 \underbrace{\text{Cheryl's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \xcancel{ \text{August 7}} & \xcancel{\text{August 9}} \\ \xcancel{\text{September 3}} & \text{September 5} & \text{September 7} & \xcancel{\text{September 10} } \\ \hline \end{array}} \qquad\underbrace{\text{Betty's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \text{June 4} & \text{June 6} & \text{June 8} & \\ \text{July 4} & \text{July 5} & \text{July 6} & \text{July 8} \\ \text{August 4} & \text{August 5} & \text{August 7} & \text{August 9} \\ \text{September 3} & \xcancel{\text{September 5}} & \text{September 7} & \text{September 10} \\ \hline \end{array}}

Following Cheryl's remark, Betty is able to determine Cheryl's birthday. This tells us that there are only one day of the month which isn't used twice, and the only possible solution to this is Cheryl’s birthday is on September 7 \boxed{\text{Cheryl's birthday is on September 7}} .

Betty then made another claim that the day of the month of her birthday is a prime number. Among the number { 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } \{3,4,5,6,7,8,9,10\} , only 3 , 5 3,5 and 7 7 are prime numbers. Thus we are able to cancel out plenty of possible dates from Betty's list:

Betty’s possible birthday dates June 4 June 6 June 8 July 4 July 5 July 6 July 8 August 4 August 5 August 7 August 9 September 3 September 5 September 7 September 10 \underbrace{\text{Betty's possible birthday dates}}_{\begin{array} { | c c c c | } \hline \xcancel{ \text{June 4} } & \xcancel{ \text{June 6} } & \xcancel{ \text{June 8} } & \\ \xcancel{ \text{July 4} } & \text{July 5} & \xcancel{ \text{July 6} } & \xcancel{ \text{July 8} } \\ \xcancel{ \text{August 4}} & \text{August 5} & \text{August 7} & \xcancel{ \text{August 9} } \\ \text{September 3} & \xcancel{\text{September 5}} & \text{September 7} & \xcancel{ \text{September 10} } \\ \hline \end{array}}

In the end, Cheryl is also able to figure out Betty's birthday. This tells us that there is only one month which isn't used twice, and the only possible solution to this is Betty’s birthday is on July 5 \boxed{\text{Betty's birthday is on July 5}} .

So the total number of days between Betty's and Cheryl's birthday (both dates inclusive) is

Total number of days in July except the first 4 days of July + total number of days in August + first 7 days of September = ( 31 4 ) + 31 + 7 = 65 . \text{Total number of days in July except the first 4 days of July} \\ + \\ \text{total number of days in August} \\ + \\ \text{first 7 days of September} \\ = \\ (31-4) + 31 + 7 =\boxed{65} .

Moderator note:

Good clear presentation of the solution, which makes it easy to follow the argument. Having such a system also makes it easier to come up with the solution.

You missed out the redundand info part.. xD But that in itself was redundant, ironically.

Irina Alexandra - 5 years, 3 months ago

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it is useful!

Joel Yip - 5 years, 3 months ago

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Well, I don't think it is, in this particular case. I mean Cheryl knows the month and that month only has one prime number left uncrossed. Even if we solved the problem leaving that out (like Pi Han Goh did) we would have gotten the correct answer.

Irina Alexandra - 5 years, 3 months ago

@Irina Alexandra You can cancel the whole September

Joel Yip - 5 years, 3 months ago

SAME way sir

Kaustubh Miglani - 5 years, 3 months ago
Tran Hieu
Feb 29, 2016

From Jul 5 to Sep 7 (both dates inclusive) we should have 65 days because July and August both have 31 days, we have 31*2 + 7-5+1 = 65

5 Jul
5 Aug 31
5 Sep 31+31=62
7 Sep 62+3=65


Joel Yip - 5 years, 3 months ago

Can you explain how you arrived at the values of Jul 5 and Sep 7?

Calvin Lin Staff - 5 years, 3 months ago

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