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Help! Can anyone solve this?

A Father Takes His Three Children To The Movies. He Is Asked How
Many Children Are Under 12. He Replies ‘my Oldest Will Or Has Never  Been Twice The Age Of The Other Two, But The Middle Child Would Have
Been Twice The Age Of The Youngest Next Year If My Youngest Was Born A
Year Later. When My Youngest Was Born, My Eldest Was One Year P;der
Than The Middle Child Now, And I’m Currently Twice The Age Of All My
Children Ages Combined. How Old Is The Father?

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Help! Can anyone solve this?

"He Replies ‘"

what are those codes?

Dragonshade

Member

Registered: 2008-01-16

Posts: 147

Re: Help! Can anyone solve this?

this kind of question is really not my thing, I guess the father would be 20+5k, k=2,4,6,8....
Father=F ,  x<y<z (3 children)

F        x      y     z
30      7      11   12
40      9      15    16
50     11      19    20

I am not sure, but man, thats one heck of a father XD

Last edited by Dragonshade (2008-04-17 03:49:57)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Help! Can anyone solve this?

"He Replies ‘"

what are those codes?

Chsnge your View Encoding to Unicode (UTF-8).

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help! Can anyone solve this?

The problem seems unsolvable.
#ZZ.)  Old > 2 Middle
#YY.)  Middle + 1 = 2 (Young-1+1)
#XX.)  Old - Young + 1 = Middle
Dad = Young + Middle + Old
Since Young is less than Middle by definition,
then #XX shows that Old is less than double middle,
however #ZZ says otherwise, so
there is no answer.

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help! Can anyone solve this?

You might consider leap day birthdays, too though.

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Help! Can anyone solve this?

Also, I was using real numbers for the
ages, not discrete integral values, so my
post # 5 is only true for real values.
This makes a difference because when
the young one is born, the old one is
infinitely older, and it descreases continuously
as time goes by, but if you go with
integral years, then the problem changes
because the "2 times" greater factor
that has never been but will be could be
passed by if not using real numbers, for
old to middle child, but not passed thru
for old to young child yet...

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igloo myrtilles fourmis

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Help! Can anyone solve this?

This might be a stretch, but the only way that I can see it working is if the middle and youngest child were born within 1 year of each other.  Here are the equations I came up with:

Middle + 1 = 2 * (Youngest - 1) = 2 * Youngest - 2
Oldest - Youngest = Middle + 1

These give us
Middle = 2 * Youngest - 3
Oldest = 3 * Youngest - 2

Now, we're subject to the fact that the oldest has never been twice the age of the younger two children.  We're also subject to the fact that the middle child has to be older than the youngest child, obvioiusly.  Here are some values based on the equations above:

Youngest    0    1    2    3    4     5
Middle       -3   -1   1    3    5     8
Oldest       -2    1    4    7   10   13

Clearly the first 3 columns are impossible.  The last 2 are also ruled out since the oldest child is or would at some point have been twice the age of the middle child.  This remains true as the ages increase, so we're only left with ages of 3, 3, and 7.  This of course depends on the youngest and middle child being born within one year of each other.  It's not common, certainly, but it is physically possible.  This would make the father 2 * (3 + 3 + 7) = 26 years old.

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Wrap it in bacon