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A few days ago, as I was delving into the mess in my grand parents' attic, I found an impressive ancient book that was written in a language that I had never seen before.

"This book is a collection of riddles printed in the eighteenth century." Grandma said. "One in particular is worth reading: it is said to be the ancestor of the age-of-the-captain riddles. Let me translate it to you."

And she proceeded:

A Captain's son asks his father:

"Why Father, I've noticed that we never celebrate your birthday, and in fact I don't think I even know when you were born."

Answers the Captain:

"If you subtract four from the day I was born, then you get a non-zero integer with at least two distinct prime divisors, one of which is the month I was born. Now with that knowledge, if I tell you the day I was born, then you shall know the year as well."

What was the Captain's date of birth? (dd/mm/yyyy)

She then closed the book and stared into my eyes with the most mischievous look I'd ever seen. She asked:

"In what language was this book written?"

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    $\begingroup$ ... is it his mother? proceeds to read beyond the title $\endgroup$
    – Riddler
    Commented Feb 11, 2019 at 22:43
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    $\begingroup$ Several means "more than two but not many", see e.g. Oxford. Using that there is no solution. $\endgroup$
    – AndyT
    Commented Feb 12, 2019 at 15:14
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    $\begingroup$ @AndyT Does it really? I've always taken it to mean "more than one". Thanks. $\endgroup$
    – Arnaud Mortier
    Commented Feb 12, 2019 at 15:31
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    $\begingroup$ Obligatory XKCD $\endgroup$
    – Steve V.
    Commented Feb 13, 2019 at 19:57
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    $\begingroup$ @ArnaudMortier - Yes, that XKCD is annoyingly on-target for native English speakers. :) $\endgroup$
    – Steve V.
    Commented Feb 13, 2019 at 20:34

4 Answers 4

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This book was written in SWEDISH.

First - the day, minus four, must have at least two prime factors ("several distinct prime divisors"). Moreover, since we will know the month uniquely, one of these primes should be greater than $12$ or the date greater than $28$ (checking $29$ and $31$: both $29-4$ and $31-4$ have only one distinct prime divisor, so neither work). Thus, the date is $2\times13+4=30$ or February 30th.

The next largest candidate would be $2\times17+4$ or February 38th, followed by February 42 and March 43rd, all of which are obviously not real dates.

So, his birthday must be on February 30th! Wait, say what? Yes, that is indeed a real thing! In particular:
    –  It satisfies the date requirement: we now know that his birthday was on February 30th, 1712
    –  We now know why we never celebrate his birthday, since February 30th never existed thereafter.
    –  And most importantly, we now know the language it was written in—namely, Swedish.

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    $\begingroup$ That’s pretty neat, but one minor nitpick - pretty sure 25 is not prime (it does have only one prime divisor though so it’s still ruled out by that) $\endgroup$
    – Jeff
    Commented Feb 11, 2019 at 23:42
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    $\begingroup$ Very good answer! $\endgroup$
    – Arnaud Mortier
    Commented Feb 12, 2019 at 8:50
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    $\begingroup$ Wow, I learned something today! $\endgroup$
    – Vincent
    Commented Feb 12, 2019 at 15:05
  • $\begingroup$ rot13("Ubj vf vg gung jr jvyy xabj uvf zbagu havdhryl?") I'm not seeing how this is a consequence from the question as stated. $\endgroup$
    – Wheat Wizard
    Commented Feb 13, 2019 at 4:15
  • $\begingroup$ W W: rot13(Vs gurer ner ng yrnfg gjb qvfgvapg cevzr snpgbef haqre guvegrra, naq gur qnl vf ng zbfg gjragl-rvtug, gura gurer ner gjb inyvq zbaguf fvapr rirel zbagu vf cerfhzrq gb unir ng yrnfg gjragl-rvtug qnlf va zbfg lrnef (whfg yvxr gur Fjrqra pnfr, fher fbzr bgure pbhagevrf rkcrevraprq fubegrarq zbaguf qhevat gur pbairefvba gb gur Tertbevna pnyraqne, ohg guvf jnf n bar-gvzr rirag.)) $\endgroup$
    – phenomist
    Commented Feb 13, 2019 at 6:40
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Since $day-4$ has multiple prime divisors and a unique date including the year can be deduced from it, there must be a weird year where some months are longer than usual so that these dates can't be seen in any other year. The only year satisfying all the conditions was 1712 in Sweden. If the given number is 26, it's Feb 30th, 1712, and the book is written in Swedish.

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  • $\begingroup$ Make sure to check other people's answers to ensure you're not posting a duplicate :D $\endgroup$
    – AHKieran
    Commented Feb 12, 2019 at 8:51
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    $\begingroup$ It's not a duplicate if you read carefully. While the accepted answer focused on the prime divisors right from the start, I mainly tried to point out that it has to make the year unique. My answer was just a lot less mathematical. $\endgroup$
    – Nautilus
    Commented Feb 12, 2019 at 9:54
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Is the answer:

02/03/1700?

Four days before that would be 26/02/1700. The month has 2 distinct prime divisors, 2 and 13, one of which is the month.

The only part of the calendar which differs from year to year is 29/02. So the 4 day stretch must span this date, if it is true that the year can be deduced by the day.

Years divisible by 100, but not 400, do not have a leap day. Of the numbers {25, 26, 27, 28, 29}, only 26 has distinct prime divisors. And 4 days ahead of 26/02 is 02/03, in a non-leap year.

The next question is the century. But the second paragraph states that the riddles were published in the 18th century (ie., the 1700s). If the Captain was alive during the 1700s then only the year 1700 makes sense.

As for the language, that has me stumped. :)

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    $\begingroup$ Hi, no unfortunately, first of all if the date was 02/03/1700 then subtracting 4 from the day would give 2-4=-2. That number doesn't have several prime factors. $\endgroup$
    – Arnaud Mortier
    Commented Feb 12, 2019 at 8:57
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Is it 30/02/1600?  30-4 = 26 (2,13 - 2 distinct prime factors)  Feb 30, so it got to be 1600 or 1200?

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    $\begingroup$ (1) Please post answers as statements, rather than questions.  (2) Please use spoiler formatting on answers.  (3) Note that two other answers have already given the same month and day as your answer.  (4) What is your basis for your statement about the year? $\endgroup$
    – Peregrine Rook
    Commented Feb 14, 2019 at 1:52

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