Difficult Brainteaser

Otis

Otis

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The hostess at her 20th wedding anniversary party tells you that the youngest of her three children likes her to pose this problem, and proceeds to explain: “I normally ask guests to determine the ages of my three children, given the sum and product of their ages. Since Smith missed the problem tonight and Jones missed it at the party two years ago, I’ll let you off the hook.”

Your response is: “No need to tell me more, their ages are…”

o_O

[imath]\;[/imath]
 
I think we are supposed to assume that each child is legitimate and that the hostess had no children by a previous marriage.
 
Last edited:
lev888 said:
What does "missed it" mean? Solved incorrectly?
Click to expand...

Hmmm, could it mean that there are two (or more) sets of integer solutions to a+b+c=<quoted sum> and abc=<quoted product> and "missed it" means that the guesser just found one of the triples that doesn't match the actual ages. For example {a1,b1,c1} and {a2,b2,c2} might be solutions to the quoted constraints but the person guessed {a1,b1,c1} when the actual ages are {a2,b2,c2}

I also guess that:-
- ages are integers
- The youngest must have been older than, say 5, two years ago (since we are told that they liked this question being posed)
- one year ago there was only one possible set of {a,b,c}
- the youngest wasn't a twin, so that a < b ≤ c

But I might be wrong, because...
I cheated and wrote a computer program but I get the oldest being 20 which would mean the first baby was born before the wedding :unsure:

My answer is ages 9, 12, and 20

There are two possible answers for this year's sum and product:- {8 15 18} {9 12 20}
Also two possible answers for two years ago:- {6 14 15} {7 10 18}
But only one possible answer for one year ago
 
I think the ages must follow these conditions and more ( couldn't think of them yet)...
1) The age of the oldest is less than 20
2) The age of the middle and youngest children are different
 
Cubist said:
Hmmm, could it mean that there are two (or more) sets of integer solutions to a+b+c=<quoted sum> and abc=<quoted product> and "missed it" means that the guesser just found one of the triples that doesn't match the actual ages. For example {a1,b1,c1} and {a2,b2,c2} might be solutions to the quoted constraints but the person guessed {a1,b1,c1} when the actual ages are {a2,b2,c2}

I also guess that:-
- ages are integers
- The youngest must have been older than, say 5, two years ago (since we are told that they liked this question being posed)
- one year ago there was only one possible set of {a,b,c}
- the youngest wasn't a twin, so that a < b ≤ c

But I might be wrong, because...
I cheated and wrote a computer program but I get the oldest being 20 which would mean the first baby was born before the wedding :unsure:

My answer is ages 9, 12, and 20

There are two possible answers for this year's sum and product:- {8 15 18} {9 12 20}
Also two possible answers for two years ago:- {6 14 15} {7 10 18}
But only one possible answer for one year ago
Click to expand...
I am obviously missing way too much here.
Why can't the youngest be a twin?
Why are there only two possibilities for this year's sum and product?
Why can't the sum be 7+8+9 =24 and the product be 7*8*9=504? Why not any 3 ages between say 5 and 20?

Edit: Now I understand everything. There are an awful lot of combinations to consider!
 
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BigBeachBanana said:
2) The age of the middle and youngest children are different
Click to expand...
Steven G said:
Why can't the youngest be a twin?
Click to expand...

Just so you know ...

I always object to puzzles based on this idea, because I am a twin, and my twin brother is recognized as the oldest in the family. A five-minute difference is not nothing. So, as demonstrated here, it is possible to talk about the "middle and youngest" even if their (integer) ages are the same.

I have made no attempt to solve this one, so I don't know whether this issue actually plays a role.
 
I missed some possible answers in post#5...
these possible answers make more sense in the context of the OP...

Ages 6, 12, 16 which gives the same sum & product as 8, 8, 18
and two years ago this would have been 4, 10, 14 which gives the same sum & product as 5, 7, 16

Also ages 8, 8, 18 which gives the same sum & product as 6, 12, 16 (see above!)
and two years ago this would have been 6 6 16 which gives the same sum & product as 4, 12, 12
 
I, too, cheated using codes as there are too many combinations to consider o_O. Personally, I wouldn't call this is a "brainteaser". Maybe someone has an elegant and methodical solution.

Assumptions:
1) The oldest child is less than 20 years old
2) The age of the two younger children are distinct integer ages (as pointed out by @Dr.Peterson above)
3) The sum and product of the ages must be the same as when Smith guessed as well as when Jones guessed them 2 years ago.

The answer is (5,6,16)--Sum= 27 ; Product = 480
Smith's guess: (4,8,15)-Sum =27 ; Product = 480
The sum of what Jones guessed 2 years ago must be 21, and the product is 168
Jones' guessed (2,7,12)
 
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BigBeachBanana said:
there are too many combinations to consider [so] I wouldn't call this is a "brainteaser".

The answer is (5,6,16)
Click to expand...
You are correct! ?

Here is the answer, as provided by the author. They don't share how they eliminated combinations.
We look for cases where (1) the oldest child is under 20, (2) the
younger two children have different ages, and (3) there is a product and
sum that give rise to ambiguity for the ages both this year and two years
ago. There is only one set of ages that accomplishes this: (5, 6, 16). The
product and sum could be achieved by (4, 8, 15), which must have been
guessed by Smith. The product and sum two years ago could be achieved
by (2, 7, 12), which must have been guessed by Jones two years ago.

It's the same author who said that finding that second taxi number is "easy".

?

[imath]\;[/imath]
 
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