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Two old school friends, Bob and Jim, meet again after many years. It's the same old town, a bottle of wine on a small wooden table in the castle garden, where the Mathematics faculty has been hidden away, untouched by time. Their wives accompany them.

  • Hey Bob, long time no see! So you're a renowned professor of history now, with a book titled The End of Wars, and politicians seeking your advice? I'm more than surprised… Oh, please, take that as a compliment!
  • To be honest, Jim, I'm surprised too. We all thought you'd get your doctorate in math before lunch back in the day. As for me, nothing much happened—just digging through books, archives, and ancient sands. I built a house, then built another one, got married young, and did that again! Now I have three beautiful daughters, and maybe that's the only thing I'm really proud of.
  • How old are they?
  • Oh, they are…

Bob pauses, smiling, and then takes out a sheet of paper. After a moment of scribbling and thinking, he continues:

  • Well, Jim, the product of their ages is exactly the year our history teacher used to call 'The Year of Fire and Fear.'
  • Oh, yes, I remember... I can almost see those digits again on the wide, tall blackboard, taking up nearly all its width and height.
  • And their sum? That's your favorite number, that we all know at this table. Can you figure it out?
  • No, Bob, I need more.
  • Alright, my youngest daughter has blue eyes.
  • Okay, that helps a lot! Do you have twins?
  • Yes! Really, Jim, you have their ages by now?!
  • Sure, Bob! I finally know something my wife doesn't!

Question: What is Jim's favorite number?

$$ \color{red}{\textbf{The text was essentially changed, please reread!}} $$ I am apologizing for this. Initially the text was giving the information $a\ne b$ for the ages $a,b,c$ (with $a\ge b\ge c$) at some intermediate point, designed to help to decide between cases. (And finally Jim's wife could also know the ages, and the point she may know this is important. And still there were some bugs.) But making sentences is not easy. So i decided to change, give the opposite information $a=b$, which is easier to combine as text to the fact that finally Jim's wife cannot decide.

Hints:

Mathematically, Jim knows initially the year $Y\ge 10$ (because it was written with digits on the blackboard, i need only $Y>0$) and his favorite number $N$, so he tries to find all triples $a,b,c$ of positive integers so that $a\ge b\ge c$, $abc=Y$, and $a+b+c=N$.

He cannot decide first, and this means that there are more solutions to this system of two diophantine equtions. Then he gets the information $b>c$ (there is a youngest daughter), and can decide, it helped a lot!

So Jim knows now the numbers, and gives us the information that $a=b$. Which should still not allow for his wife to decide, because he claims it so!

With the last information, we can find $Y,N;a,b,c$.

Here is a toy example suggesting what works and what not - inspired from the comments of Tim Seifert. (I am very thankful for the many comments!) Assume for a moment that $N=14$ is the favorite number of Jim. We are considering the story from the pespective of Jim's wife. This is a particular value that has exactly two possible years and corresponding matching solutions $(a,b,c)$, the two triples are displayed in the same line with $Y,N$ below... $$ \begin{aligned} Y&= 40\ , & N&= 14\ , &&(8,5,1)\ ,\quad(10,2,2)\ ,\\ Y&= 72\ , & N&= 14\ , &&(6,6,2)\ ,\quad(8,3,3)\ . \end{aligned} $$ There must be (at least) two solutions $(a,b,c)$ for a given pair $(N,Y)$, with $a\ge b\ge c$, $abc=Y$, $a+b+c=N$, else Jim decides immediately. Also, all but one solutions must have $b=c$, else there is no use of the information $b\ne c$ (the blue eyes), and Jim cannot eliminate any triple. After having this information Jim can decide. There is a postlude, Jim asks if Bob has twins. Yes, so it is. In the above case this matches only the case $Y=72$ and $(a,b,c)=(6,6,2)$. But then Jim's wife can also decide... So $N$ is not $14$...

P.S. I hope you'll enjoy the joke! This is a puzzle I composed myself. Just to clarify, Jim's wife didn't know anything about the other numbers involved at the beginning of the conversation, she knows only the favorite number. Also, to be precise, "same age" means they are exactly the same age (as numbers), not older. If two daughters were born in the same year, we assume they are twins, and the few minutes between their births do not make one twin (slightly) older.

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  • $\begingroup$ I'm probably missing something here, but how can we (or Jim's wife!) possibly tell? For instance the two favorite numbers 3 and and 14 (with the initial ambiguity given by (0,0,3) vs. (0,1,2) and (2,2,10) vs. (1,5,8) respectively) both seem entirely possible. (These are just the first examples that I found - I'm positive that there are more with nonzero product, if that is supposed to be an implicit constraint). $\endgroup$
    – Tim Seifert
    Commented Aug 14 at 11:23
  • $\begingroup$ @TimSeifert Oh, it is my fault that zero is also allowed so far, i will edit to avoid this number 0 (for the product), and add some more hints. $\endgroup$
    – dan_fulea
    Commented Aug 14 at 11:35
  • $\begingroup$ @TimSeifert I am so sorry, an essential detail was left out in the process of translation, it is imperiously necessary to mention that Jim's wife knows his favorite number. English is not my native language, translation tools, that sentence was lost in the process... Thanks for the early attempt, please apologize the early errors. Hope it is all fine by now. $\endgroup$
    – dan_fulea
    Commented Aug 14 at 12:12
  • $\begingroup$ @dan_fulea Zero is already ruled out as any of the daughters' ages, since it would make the product zero - but there was no Year 0. $\endgroup$
    – Nuclear Hoagie
    Commented Aug 14 at 13:06
  • $\begingroup$ @NuclearHoagie The work done to have the number theoretical situation above was relatively huge, so my focus was and is after that to write a clear and unambiguous, by no means misleading story. Since i never had a zero in mind all the time, i completely forgot about the possibility that somebody may consider also this case, Tim Seifert mentioned this case in his comment above, and i am thankful and highly appreciate his words. This is no trick question, so to be sure nobody would suspect something "unusual", i ruled out by grammar (digits) any chance of considering a zero year. $\endgroup$
    – dan_fulea
    Commented Aug 14 at 13:23

1 Answer 1

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Not the way it's supposed to be solved, but I guess the number is...

42. There's an extremely popular meme portraying this number as the answer to everything.

It could be $x + x + y$ or $z + t + t$, with the possible ages following $x<y$ and $z<t$. When I fed the equation $(42-2x)x^2 = (42-2t)t^2$ to WolframAlpha, I got $x=5, t=20$ and $x=9, t=18$. I doubt $5*5*32 = 800$ has anything to do with "fire and fear", so $9*9*24=1944$ might be a better answer, coming just 1 year before the end of WW2.

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    $\begingroup$ Oh, yes, this is the answer to the question, presumably to all questions following a general opinion - although some people have doubts. At any rate this question does not dance against the general opinion. But why does it work? There are although some more details to cover, for instance why could Jim not decide earlier, and if there are some other possibilities... Surely, i am impressed, excellent work or excellent guess! $\endgroup$
    – dan_fulea
    Commented Aug 15 at 11:13

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