Triplets
Philip Carter
Is there any possible way of solving the following puzzle by formula?
A woman has 7 children. On multiplying their ages together she obtains
the number 6591.
Given that today is the birthday of all 7, how many sets of triplets are
there, and what are the ages of all 7 children?
Philip Carter
Visit: The Enigmatic World of Philip Carter
http://www.knowl.demon.co.uk
GimmeFuel
>
>A woman has 7 children. On multiplying their ages together she obtains
>the number 6591.
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Not sure if "by formula" you mean algebraically or just not by
guessing and checking, but here's how I did it:
Since today is all of their birthdays, their ages are all integers. So
there must be 7 integers which multiply together to make 6591. If you
take the prime factorization of 6591, you get 3 * 13 * 13 * 13. Having
one year old triplets won't change the product of their ages, so you
end up with 7 kids: a set of 1 year old triplets, a 3 year old, and a
set of 13 year old triplets.
--
if(democrats == republicans) {vote(libertarian);}
grapheus
> >Is there any possible way of solving the following puzzle by formula?
> >
> >A woman has 7 children. On multiplying their ages together she obtains
> >the number 6591.
>
>
> guessing and checking, but here's how I did it:
>
> Since today is all of their birthdays, their ages are all integers. So
> there must be 7 integers which multiply together to make 6591. If you
> take the prime factorization of 6591, you get 3 * 13 * 13 * 13. Having
> one year old triplets won't change the product of their ages, so you
> end up with 7 kids: a set of 1 year old triplets, a 3 year old, and a
> set of 13 year old triplets.
On a purely mathematical point of view, this is not the only solution
of the problem. For instance, she could have no triplet at all, with
4 children of 1 year, two of 13 years and one of 39 years. But is this
plausible ?.. Can a woman of minimum 39+18 = 57 years be to-day the
mother of quadruplets with the progress of medecine ?...
I believe to avoid this second solution, it should be said that "she
has at least one triplet".
grapheus