Once upon a time, there were 3 charming sisters, who attended the luxurious banquet in the royal palace. The Prince, stunned by their fascinating beauty, approached his 3 new guests, and after brief introduction, he politely asked them about their ages.

Then the 3 ladies, named A, B, C, whispered to one another before coming up with a little plan to test the Prince's wit.

Lady C: The age sum of Lady A and Lady B has a remainder of 3 when it is divided by 5. And only one of us has an age as multiple of 5.

Lady A: The age sum of Lady B and Lady C has a remainder of 5 when it is divided by 6. And only one of us has an age as multiple of 6.

Lady B: The age sum of Lady C and Lady A has a remainder of 4 when it is divided by 7. And only one of us has an age as multiple of 7.

Prince: How clever of you! Judging from my sight, Lady A is the youngest, Lady B the middle, and Lady C the eldest, I bet.

3 ladies: Correct! Your Highness! The least composite numbers applied are indeed our ages!

Prince: Thank you, fair ladies. Now I know all of your ages.

What is the least possible sum of the 3 sisters' ages?

The answer is 59.

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Oh, well, I just put "least possible" sum to make sure they are not fairies in disguise. :)

Worranat Pakornrat
- 3 years ago

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Ahh, I was just wondering when the next solution would crop up. A long ways off. I found the sisters' ages by trial and error, looking at all the multiples of 5 between 10 and 100, and then finding nearest multiples of 6 and 7. Process of elimination.

Michael Mendrin
- 3 years ago

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Oh, can't we use multivariate Chinese remainder theorem for this?

Worranat Pakornrat
- 3 years ago

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@Worranat Pakornrat – too much work... I figured that there weren't too many triples that I needed to examine

Michael Mendrin
- 3 years ago

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@Michael Mendrin – I see. Thus, I adjusted the dialogue a bit, so one would find these sisters more mysterious. :)

Worranat Pakornrat
- 3 years ago

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@Worranat Pakornrat – Oh oh... well now

Michael Mendrin
- 3 years ago

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@Michael Mendrin – Maybe you better think twice about that? It doesn't seem like {18,20,21} would be an unique solution {18,25, 28} would be another. I suggest you go back to the original wording.

Michael Mendrin
- 3 years ago

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@Michael Mendrin – I agree it's not unique, but I still think the solution is the least possible sum.

Worranat Pakornrat
- 3 years ago

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@Worranat Pakornrat – It's the least possible sum only if none of them are preteens.

Michael Mendrin
- 3 years ago

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@Michael Mendrin – Well, we still have the 5,6,7 multiple constraints and A being the youngest. No lesser triple could apply.

Worranat Pakornrat
- 3 years ago

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@Worranat Pakornrat – How about 12, 20, 21?

Michael Mendrin
- 3 years ago

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@Michael Mendrin – 12+20 = 32 = 2 (mod 5). It doesn't fit their first rhyme 3 mod 5.

Worranat Pakornrat
- 3 years ago

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@Worranat Pakornrat – Who says that b is the next youngest? 12+21=33=3(mod 5). Maybe fix your wording a bit more carefully, and then you have it?

Michael Mendrin
- 3 years ago

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@Michael Mendrin – OK. Now the order is specified.

Worranat Pakornrat
- 3 years ago

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@Worranat Pakornrat – Now the problem is harder

Michael Mendrin
- 3 years ago

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@Michael Mendrin – Time to use multivariate CRT now. :)

Worranat Pakornrat
- 3 years ago

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Or the sisters could be $228, 230, 231$

If $\;\; x= 2 \cdot 3 \cdot 5 \cdot 7 \cdot n \;\;$ where $n$ is an integer

$(18+x, 20+x, 21+x) \;\;$ is the general solution