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Louis Thépault offers this puzzle in his book Le chat à six pattes et autres casse-tête : 100 petits problèmes de mathématiques très amusants.

Four men, former classmates, met in a restaurant, each accompanied by his wife. Here’s what we know about these eight people:

  1. Alexandre and his wife were born in the same year, both on Mondays. In the year of their wedding Alexandre’s birthday fell on a Friday, and his wife’s on a Thursday.

  2. Noëlle, who was given that name because she was born on the 25th of December, is older than her husband.

  3. Clara was born in 1982, and her husband was born one January morning.

  4. Justine arranged her wedding to fall on the day prior to her birthday, and Dominique vice versa: the wedding on the next day after the birthday; moreover, exactly six days passed between these two weddings: Justine’s was on a Friday and Dominique’s on the next Thursday.

  5. Claude, Nicolas’ wife, and Frédéric were born in the same year on Fridays the 13th, Frédéric being the youngest.

  6. No two people share a birthday.

  • Who married whom?
  • Whose wedding was on the 12th of May, 2005, the day before their birthday?
  • Name the exact date of Clara’s wedding.

Classmates means the ages of all eight people don’t vary greatly. Each of the four men was married exactly once to exactly one of the four women, nobody changed their sex/gender etc. All other confusing things are confusing on purpose :-)

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  • $\begingroup$ I'm assuming "year" means calendar year. Is that not correct? $\endgroup$ – Chris Sunami supports Monica Oct 23 '20 at 16:17
  • $\begingroup$ @ChrisSunamisupportsMonica Well yes, the usual Gregorian year, what else could it be? $\endgroup$ – Roman Odaisky Oct 23 '20 at 16:35
  • $\begingroup$ I thought maybe it meant "within a span of 365 days." If this isn't playing off ambiguity in the term "year" then I'm with @HRogers, none of the ladies can be Alex's bride. $\endgroup$ – Chris Sunami supports Monica Oct 23 '20 at 18:00
  • $\begingroup$ Having gone and looked up the book referenced, I can confirm that the puzzle does have a valid solution as stated. It requires a good deal of lateral thinking though. I'll post the full answer in a couple weeks if nobody has cracked it by then. $\endgroup$ – H Rogers Oct 23 '20 at 18:33
  • $\begingroup$ @HRogers It requires absolutely no lateral thinking. What it requires is being very careful with the assumptions. $\endgroup$ – Roman Odaisky Oct 23 '20 at 22:01
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I believe the twist here is that:

Dominique was born on Dec 31st, or Justine was born on Jan 1st.

Why do we need the twist?

It's not too hard to rule out everyone else as being Alexandre's wife. Nicolas is definitely a man, and Claude and Frederic (likely men anyway) were born on Fridays. Noelle being older than her husband doesn't fit with being born in the same year, both on Mondays. Clara being born in 1982 means if she was Alexandre's wife they'd have had to be married in a leap year and he couldn't have been born in January.

From there

Consider Dominique being born on Dec 31st. Then her wedding was on a Thursday, January 1st. Alexandre's birthday could be any Friday in January or February, and then Dominique's birthday would be Thursday, Dec 31st of that year. This requires that they were born on a leap year and married in a non-leap year. Because Clara was born in 1982, Dominique and Alexandre must've been born in either 1980 or 1984 to satisfy the ages not varying greatly. Dec 31st 1980 was a Wednesday, but Dec 31st, 1984 was a Monday. Being born in 1984, they could have married in either 2009 or 2015, as those are non-leap years with Dec 31st falling on a Thursday.

Now consider Justine being born on Jan 1st. Then her wedding was Friday, Dec 31st, and her birthday was Saturday Jan 1st of the next year. So Alexandre and Justine must have been married on a leap year, and Justine's birthday in the year of their wedding was Thursday Jan 1st. However, if you look at Jan 1sts between 1978 and 1986 you'll find that only 1979 had Jan 1st as a Monday. They could have then have been married in 2004, which is a leap year in which Jan 1st was a Thursday.

Now that we have some possibilities established for that person, let's move on to another one.

Let's figure out who Nicolas' wife is. His wife was born on a Friday 13th, so Noelle, Justine, and Dominique are out. The last three people are Claude, Frederic, and Clara. Clue five appears to state that three people were born on three different Friday 13ths in a single year. However, in 1982, when Clara was born, there was only one Friday 13th. So the clue should instead be interpreted as saying "Claude, who is Nicolas' wife, and Frederick..." This results in the women being Noelle, Justine, Clara, and Claude, which also explains why the wording in clue four is so careful to not use a pronoun for Dominique.

We now know who is married to whom:

Alexandre is married to Justine. They were born in 1979, and married on December 31st 2004. Nicolas is married to Claude. Clara married Dominique on January 6th, 2005 - the only Friday, Jan 13ths near 1979 were in 1978 and 1984, so to be born in January Frederic would either need to be about 5 years younger (not classmates) or the oldest. This leaves Frederic and Noelle being married.

As for the May 12th, 2005 wedding

We don't yet know when Nicolas and Claude or Frederic and Noelle were married. From Wikipedia we can see that when May 13th is a Friday it is the only one in the year, so neither Claude nor Frederic were born in May. This leaves Nicolas as the only possibility, so Nicolas and Claude were married on May 12th, 2005 before Nicolas' birthday the next day.

When was Clara's wedding?

January 6th, 2005

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  • $\begingroup$ This contradicts the first part of statement 4 unfortunately $\endgroup$ – Roman Odaisky Oct 27 '20 at 14:46
  • $\begingroup$ This is for sure a breakthrough though! If you take a good look at Joe Ferndz's answer you probably have enough to solve it. $\endgroup$ – H Rogers Oct 27 '20 at 14:49
  • $\begingroup$ @RomanOdaisky I was thinking there was some problem (especially with how it resulted in the question about the May wedding and Clara's wedding date being redundant), but I was already up too late to bother going through and double checking my work. Did I get it this time? $\endgroup$ – Rob Watts Oct 27 '20 at 17:56
  • $\begingroup$ @RobWatts congrats, you should probably edit the first part of your response to reflect the final answer. $\endgroup$ – H Rogers Oct 27 '20 at 20:46
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Edit: See comment above. There is a correct solution to the puzzle. Leaving my original (incomplete, incorrect) response for reference.


I may be wrong, but it appears that this puzzle has no correct solution. The easiest way to demonstrate this is to prove that Alexandre cannot be married to any of the four women.

Noelle: Noelle is born on December 25th and is older than her husband. Since Alexandre and his wife were both born on Mondays in the same year, and there is no Monday that comes in a given year following Monday December 25th, Alexandre and Noelle cannot be married.

Justine and Dominique: In the year of Alexandre's wedding his wife's birthday falls on a Thursday, however in the year of Justine's wedding Justine's birthday falls on a Saturday (day after her wedding) and in the year of Dominique's wedding Dominique's birthday falls on a Wednesday (day before her wedding). Therefore, Alexandre cannot be married to either Justine or Dominique.

Clara: The only way for two given dates to land on the same day of the week in one year and a different day of the week in another year (as Alexandre and his wife's birthdays are in their years of birth/marriage) is for one of the years in question to be a leap year. The dates in question must also straddle the end of February. Since 1982 is not a leap year, then the only option is that Clara and Alexandre got married in a leap year. If Clara's birthday came before the leap day and Alexandre's birthday came after, then it would be possible for Alexandre's birthday to be on Friday and Clara's to be on Thursday in their wedding year. However, we are told that Clara's husband was born in January, therefore it is not possible for their birthdays to fall on the given days of the week in their wedding year. Alexandre cannot be married to Clara.

It's possible there's some obscure thing about a day of the week vanishing somewhere in the 80s that I'm not aware of, but barring that the puzzle appears impossible. Or alternatively there's some small mistake in the phrasing (maybe Alexandre's birthday was on Thursday in his wedding year and his wife's was on Friday?).

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  • $\begingroup$ No, this apparent contradiction does not follow from the statement of the puzzle. $\endgroup$ – Roman Odaisky Oct 23 '20 at 14:24
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    $\begingroup$ Any chance you could give a hint as to where I've made a mistake? $\endgroup$ – H Rogers Oct 23 '20 at 14:53
  • $\begingroup$ I think you got Justine and Dominique's birthdays mixed up with their wedding days ... the wedding on the next day after the birthday; moreover, exactly six days passed between these two weddings: Justine’s was on a Friday and Dominique’s on the next Thursday. This is talking about their wedding days not their birth days. Thats my understanding. So Justine’s birthday on her wedding is Thursday. I paired Alexandre and Justine as husband and wife. The rest I am working thru. $\endgroup$ – Joe Ferndz Oct 23 '20 at 15:21
  • $\begingroup$ @JoeFerndz The six days in question: Friday, Justine’s wedding; Saturday, Justine’s birthday; Sunday; Monday; Wednesday, Dominique’s birthday; Thursday, Dominique’s wedding. $\endgroup$ – Roman Odaisky Oct 23 '20 at 15:39
  • $\begingroup$ @RomanOdaisky, you are right. I got mixed up. I did the reverse. Instead of using Saturday, I did Thursday. One day prior to the birthday got mixed up with wedding day. So Justine is not Alexandra's wife. Got it. $\endgroup$ – Joe Ferndz Oct 23 '20 at 17:02

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