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NEWS | Aug. 18, 2015

August 2015 Puzzle Periodical

By Roger B., Cryptanalytic Mathematician, NSA

 

Puzzle created by Roger B., Cryptanalytic Mathematician, NSA

Nadine is having a party, and she has invited three friends – Aaron, Doug, and Maura. The three of them make the following statements on the days leading up to the party:

Two Days Before the Party:
Aaron: Doug is going to the party.
Doug: Maura is not going to the party.
Maura: Aaron will go to the party if, and only if, I do.

The Day Before the Party:
Aaron: Maura will go to the party if, and only if, I don't go.
Doug: An even number out of the three of us are going to the party.
Maura: Aaron is going to the party.

The Day of the Party:
Aaron: It is not yet 2018.
Doug: Aaron will go to the party if, and only if, I do.
Maura: At least one of the three of us is not going to the party.

Nadine also knows that out of Aaron, Doug, and Maura:
One of them never lies.
A different one of them lies on days of the month that are divisible by 2, but is otherwise truthful.
The remaining one of them lies on days of the month that are divisible by 3, but is otherwise truthful.

(1) Can you figure out who is going to attend? 
(2) Can you figure out on what day, month, and year the party will be held, assuming it takes place in the future?

Click to see the answer!

(1) Attendees solution:

The first step is to realize that the date rules imply that nobody may lie on two consecutive days.

If Maura's third statement is false, then everyone is going to the party and Doug's first two statements are lies, which is impossible. Therefore Maura's third statement is true, and at least one person is not going to the party.

If Doug's second statement is false, then (from the above) exactly one person is going to the party. This can't happen without making Doug's first or third statement also false, which would have Doug lying on two consecutive days—an impossibility. Therefore Doug's second statement is true, and there are an even number of people going to the party.

If Doug is not going to the party, then Aaron's first statement is false, so his second statement must be true. This would result in an odd number of people at the party, which we have shown is not the case. Therefore Doug is going to the party.

Since an even number of people are going to the party, only one of Aaron or Maura is going, making Maura's first statement false. Therefore her second statement must be true, and Aaron is going to the party while Maura is not. The attendees are Doug and Aaron.

(2) Date solution:

Having established the attendees as Doug and Aaron, now we know that the only lies are Maura's first statement and potentially Aaron's third statement.

If the three days do not cross a month boundary, then either the second date or both the first and third dates would be divisible by 2, but nobody is available to lie with that pattern. Therefore either the second or third day is the first of a month.

If the day before the party was the first of a month, then the day of the party is the second of a month; Aaron would have to be the one who lies on dates divisible by 2. Then, for Maura but not Aaron to lie on the first day, it would have to be divisible by 3 but not 2. This is never true of the last day of a month. Therefore, the day before the party is not the first of a month, so the day of the party itself must be the first day of a month.

This makes the day before the party the last day of a month, and since nobody lies on that day it must not be divisible by 2.

This means two days before the party, the date is divisible by 2, so it must not also be divisible by 3 or there would be two liars on that day. The only way this can happen two days before the end of a month is when that day is February 28 of a leap year.

Since nobody lies on the first of a month, Aaron's third statement is true and it is not yet 2018.

Finally, since the only leap year before 2018 is 2016, we conclude that the party is being held on March 1, 2016.