This week’s brain teaser is a logic puzzle that has gained  a fair bit of fame, as it was featured as one of the tasks in the Singapore and Asian School Math Olympiads. Here goes:

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

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

Albert: I don’t know when Cheryl’s birthday is, and I know that Bernard does not know either.

Bernard: At first I didn’t know when Cheryl’s birthday was, but now I do.

Albert: Then I also know when Cheryl’s birthday is.

Last week, we met John and Seamus, but how many cows do they each have? John has seven, and Seamus five. How did I arrive at that? Let’s see. First, we know that the difference between the two numbers must be very small, as a shift of one increment can both double and completely remove the difference. We also know that the numbers must be odd, because the end numbers must be even. Armed with that knowledge, we can brute force our way to the solution:

If John were to have three cows, and Seamus one, Seamus would be left with no cows at all when giving his one cow to him.

If John had five cows, and Seamus three, the results would not match up; six is three times two.

If John had seven cows, and Seamus five, the results do match up. Seven plus one is eight, which is twice as much as four. Seven minus one is six.

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