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riddles >> hard >> Ingram's Family
(Message started by: K Sengupta on Feb 3rd, 2006, 9:53pm)

Title: Ingram's Family
Post by K Sengupta on Feb 3rd, 2006, 9:53pm
I am looking for an Algebraic Solution to the under noted  problem  which I found in a puzzle book.

PROBLEM:

The product of the ages of Ingram’s children is the square of the sum of their ages. Ingram has less than eight children . None of his children are more than 14 years old. All  the ages of his children are different.  All of his children are at least two years old.  All the ages are expressed in  integral number of years. How many children are there? What are their ages?

Title: Re: Ingram's Family
Post by JocK on Feb 4th, 2006, 1:24am
A straightforward solution based on [hide](8*2)^2 = 2^8[/hide] is: [hide]seven children aged 4, 2, 2, 2, 2, 2, and 2[/hide].



Title: Re: Ingram's Family
Post by fatball on Feb 4th, 2006, 11:24am
I have found 5 more solutions:
[hide]6 children: 4,4,2,2,2,2
5 children: 12,6,2,2,2
4 children: 4,4,4,4
3 children: 9,9,9
No solution for 2 children or less.[/hide]

Title: Re: Ingram's Family
Post by pex on Feb 4th, 2006, 12:22pm
For what it's worth:
[hide]5 children: 2, 2, 4, 4, 4
4 children: 2, 4, 6, 12
4 children: 2, 5, 5, 8
4 children: 3, 3, 6, 6[/hide]

Title: Re: Ingram's Family
Post by JohanC on Feb 4th, 2006, 2:11pm

on 02/04/06 at 12:22:03, pex wrote:
[hide]4 children: 2, 4, 6, 12
[/hide]

Up till now, you encountered the only answer with all ages different. Maybe that was the original puzzle's intention?

Title: Re: Ingram's Family
Post by JocK on Feb 4th, 2006, 5:11pm

on 02/04/06 at 14:11:06, JohanC wrote:
Up till now, you encountered the only answer with all ages different. Maybe that was the original puzzle's intention?


Yep, the riddle requires some additional constraint, something like: "the Ingram's have no twins and all children are their own natural children"...


Title: Re: Ingram's Family
Post by K Sengupta on Feb 10th, 2006, 11:32pm
Necessary  amendments incorporated in the problem body.



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