My Three Sons (and their ages)

[bonanova]

At the annual cinco de mayo Brain Den party a few days ago
I was joined by plasmid, k-man and Yoruichi-san, among others.
I happened to mention to Y-san that the ages of my three children

make for a nice puzzle.

I ask for their ages after giving their sum and product, I said.

Y-san said the puzzle was well known, and she seemed surprised

that I would use it.

It may not be that easy, I replied.

In fact, at this very party four years ago I gave the challenge to k-man,
along with the information that my oldest child had not yet turned 17,
and he was unable to solve the problem.

With as much respect as I have for k-man, Y-san replied,
I don't think that means the puzzle is unsolvable.

But just this evening I posed the question to plasmid,
adding that my oldest child had not yet turned 21,
and he too was unable to give me their ages.

Well do any two of the ages differ by 1? she asked.

After I answered, Y-san took my napkin and,
with a wink and a smile that only she knows,
wrote down the three ages.

What are they?

[k-man]

My reasoning was correct, but I just made a silly mistake at the end.

BTW, I was just kidding about getting respect, but in my solution I was still assuming that the puzzle was indeed unsolvable 4 years ago.

The ages are 18,16 and 10.

Most combinations of 3 ages will result in a distinct sum and product making the puzzle solvable given the sum and the product of the ages. I found a total of 88 sets of 3 ages (in the range 1-20) that result in non-unique sum and product. These 88 sets of ages make up 44 set pairs with 44 unique sum and product combinations:

sum, product:    age set 1;    age set 2;
 13,      36:    6,  6,  1;    9,  2,  2;
 14,      40:    8,  5,  1;   10,  2,  2;
 14,      72:    6,  6,  2;    8,  3,  3;
 16,      90:    9,  5,  2;   10,  3,  3;
 17,     144:    8,  6,  3;    9,  4,  4;
 19,     144:    9,  8,  2;   12,  4,  3;
 20,      90:   10,  9,  1;   15,  3,  2;
 21,      96:   12,  8,  1;   16,  3,  2;
 21,     168:   12,  7,  2;   14,  4,  3;
 21,     240:   10,  8,  3;   12,  5,  4;
 22,     360:    9,  8,  5;   10,  6,  6;
 23,     360:   10,  9,  4;   12,  6,  5;
 24,     240:   12, 10,  2;   16,  5,  3;
 25,     360:   12, 10,  3;   15,  6,  4;
 26,     270:   15,  9,  2;   18,  5,  3;
 26,     288:   12, 12,  2;   18,  4,  4;
 27,     480:   15,  8,  4;   16,  6,  5;
 28,     320:   16, 10,  2;   20,  4,  4;
 28,     432:   16,  9,  3;   18,  6,  4;
 28,     450:   15, 10,  3;   18,  5,  5;
 28,     560:   14, 10,  4;   16,  7,  5;
 28,     576:   12, 12,  4;   16,  6,  6;
 28,     630:   14,  9,  5;   15,  7,  6;
 29,     360:   15, 12,  2;   20,  6,  3;
 29,     504:   14, 12,  3;   18,  7,  4;
 29,     720:   12, 12,  5;   15,  8,  6;
 30,     840:   14, 10,  6;   15,  8,  7;
 31,     720:   15, 12,  4;   18,  8,  5;
 31,    1008:   12, 12,  7;   14,  9,  8;
 32,     720:   18, 10,  4;   20,  6,  6;
 32,    1008:   14, 12,  6;   16,  9,  7;
 33,     840:   15, 14,  4;   20,  7,  6;
 34,     900:   15, 15,  4;   20,  9,  5;
 34,    1152:   16, 12,  6;   18,  8,  8;
 35,    1080:   18, 12,  5;   20,  9,  6;
 35,    1120:   16, 14,  5;   20,  8,  7;
 35,    1260:   15, 14,  6;   18, 10,  7;
 35,    1440:   15, 12,  8;   16, 10,  9;
 36,    1200:   16, 15,  5;   20, 10,  6;
 37,    1440:   16, 15,  6;   20,  9,  8;
 38,    1800:   15, 15,  8;   18, 10, 10;
 40,    2160:   16, 15,  9;   18, 12, 10;
 41,    2160:   18, 15,  8;   20, 12,  9;
 44,    2880:   18, 16, 10;   20, 12, 12;

If we subtract 4 from these sets of ages we will get the ages 4 years ago and most of these sets would result in a solvable puzzle 4 years ago. The only 3 sets of ages that would make this puzzle also unsolvable 4 years ago are these:

sum, product:     ages now;  4 years ago;
 33,     840:   20,  7,  6;    16,  3, 2;
 35,    1440:   16, 10,  9;    12,  6, 5;
 44,    2880:   18, 16, 10;    14, 12, 6;

Two of the 3 sets have 1 year age difference, so if the answer to the last question was "yes" the puzzle would still not have a unique answer. The only unique answer is possible if the answer was "no".

[Pickett]

hmm...so far I have it down to 25 possibilities...still working on it...but probably won't get to it before tomorrow, at which point this will probably be solved. Great variation on this type of puzzle though!! Thanks Bonanova!

[k-man]

It's hard to get some respect around here

Their ages are 16, 10 and 9.

Thanks for a great puzzle and an honorable mention, Bonanova!

[bonanova]

Pickett: There is still time ...

k-man: thanks for being a sport - and you are well respected. But, read the ending carefully.

[Pickett]

My reasoning was correct, but I just made a silly mistake at the end.

BTW, I was just kidding about getting respect, but in my solution I was still assuming that the puzzle was indeed unsolvable 4 years ago.

The ages are 18,16 and 10.

Most combinations of 3 ages will result in a distinct sum and product making the puzzle solvable given the sum and the product of the ages. I found a total of 88 sets of 3 ages (in the range 1-20) that result in non-unique sum and product. These 88 sets of ages make up 44 set pairs with 44 unique sum and product combinations:

sum, product:    age set 1;    age set 2;
 13,      36:    6,  6,  1;    9,  2,  2;
 14,      40:    8,  5,  1;   10,  2,  2;
 14,      72:    6,  6,  2;    8,  3,  3;
 16,      90:    9,  5,  2;   10,  3,  3;
 17,     144:    8,  6,  3;    9,  4,  4;
 19,     144:    9,  8,  2;   12,  4,  3;
 20,      90:   10,  9,  1;   15,  3,  2;
 21,      96:   12,  8,  1;   16,  3,  2;
 21,     168:   12,  7,  2;   14,  4,  3;
 21,     240:   10,  8,  3;   12,  5,  4;
 22,     360:    9,  8,  5;   10,  6,  6;
 23,     360:   10,  9,  4;   12,  6,  5;
 24,     240:   12, 10,  2;   16,  5,  3;
 25,     360:   12, 10,  3;   15,  6,  4;
 26,     270:   15,  9,  2;   18,  5,  3;
 26,     288:   12, 12,  2;   18,  4,  4;
 27,     480:   15,  8,  4;   16,  6,  5;
 28,     320:   16, 10,  2;   20,  4,  4;
 28,     432:   16,  9,  3;   18,  6,  4;
 28,     450:   15, 10,  3;   18,  5,  5;
 28,     560:   14, 10,  4;   16,  7,  5;
 28,     576:   12, 12,  4;   16,  6,  6;
 28,     630:   14,  9,  5;   15,  7,  6;
 29,     360:   15, 12,  2;   20,  6,  3;
 29,     504:   14, 12,  3;   18,  7,  4;
 29,     720:   12, 12,  5;   15,  8,  6;
 30,     840:   14, 10,  6;   15,  8,  7;
 31,     720:   15, 12,  4;   18,  8,  5;
 31,    1008:   12, 12,  7;   14,  9,  8;
 32,     720:   18, 10,  4;   20,  6,  6;
 32,    1008:   14, 12,  6;   16,  9,  7;
 33,     840:   15, 14,  4;   20,  7,  6;
 34,     900:   15, 15,  4;   20,  9,  5;
 34,    1152:   16, 12,  6;   18,  8,  8;
 35,    1080:   18, 12,  5;   20,  9,  6;
 35,    1120:   16, 14,  5;   20,  8,  7;
 35,    1260:   15, 14,  6;   18, 10,  7;
 35,    1440:   15, 12,  8;   16, 10,  9;
 36,    1200:   16, 15,  5;   20, 10,  6;
 37,    1440:   16, 15,  6;   20,  9,  8;
 38,    1800:   15, 15,  8;   18, 10, 10;
 40,    2160:   16, 15,  9;   18, 12, 10;
 41,    2160:   18, 15,  8;   20, 12,  9;
 44,    2880:   18, 16, 10;   20, 12, 12;

If we subtract 4 from these sets of ages we will get the ages 4 years ago and most of these sets would result in a solvable puzzle 4 years ago. The only 3 sets of ages that would make this puzzle also unsolvable 4 years ago are these:

sum, product:     ages now;  4 years ago;
 33,     840:   20,  7,  6;    16,  3, 2;
 35,    1440:   16, 10,  9;    12,  6, 5;
 44,    2880:   18, 16, 10;    14, 12, 6;

Two of the 3 sets have 1 year age difference, so if the answer to the last question was "yes" the puzzle would still not have a unique answer. The only unique answer is possible if the answer was "no".

Hmm...I must have a flaw in my program as this was the approach I was going with, but I had more than the 44 sets that you came up with (which is why I couldn't narrow it down at the end)...nice solve k-man...I'll have to go back and see where my program is wrong.