Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Ages of Three Children

Date: 9/4/96 at 20:40:5
From: Jason
Subject: Ages of Three Children

During a recent census, a man told the census taker that he had three  
children. When asked their ages, he replied, "The product of their 
ages is 72. The sum of their ages is the same as my house number."
  
The census taker ran to the door and looked at the house number.  
"I still can't tell," she complained.  The man replied, "Oh that's 
right,  I forgot to tell you that the oldest one likes chocolate 
pudding."  

The census taker promptly wrote down the ages of the 3 children.  How 
old are they?


Date: 9/5/96 at 13:47:14
From: Doctor Tom
Subject: Re: Ages of Three Children

Hi Jason,

After I know that the ages multiply to 72, here is a complete
list of the possibilities:

Ages:            Sum of ages:
1 1 72            74
1 2 36            39
1 3 24            28
1 4 18            23
1 6 12            19
1 8 9             18
2 2 18            22
2 3 12            17
2 4 9             15
2 6 6             14  **
3 3 8             14  **
3 4 6             13

Note that every combination of possible ages which has a product of 72 
has its own unique sum of ages - except for 2, 6, 6 and 3, 3, 8, both 
of which share the sum of 14. Since the census taker can't figure out 
the ages after looking at the house number, the house number must be 
14, because then the ages could be either 2, 6, 6 or 3, 3, 8.

Now, the next clue is that the _oldest_ child likes chocolate pudding. 
This means that there is _one_ oldest child. Well, there is no oldest 
child of the ages are 2, 6, 6, so the ages of the children must be 3, 
3, and 8 years old.

-Doctors Tom and Chuck,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   


Date: 11/22/2008 at 17:22:11
From: John
Subject: The census taker logic problem

The fact the you have an oldest child does not mean there cannot 
be two children that same age.  Therefore logically the problem
cannot be solved.  I have twin daughters and one is 3 minutes older
than the other.


Date: 11/22/2008 at 23:40:43
From: Doctor Peterson
Subject: Re: The census taker logic problem

Hi, John.

I've made the same comment more than once, because I'm in that
position myself.  Dr. Rick (another Math Doctor) is my twin brother,
five minutes older than I am, and I've been aware all my life that he
is the oldest!

It's not quite as bad as you say, though; in real life, though we
shouldn't just assume anything, we can look at the facts and figure
out what someone else is thinking, rightly or wrongly.  In this case,
once you see that the two remaining possibilities are distinguished by
the presence of twins, you can see what is intended even if you know
it isn't quite right.  This is not a useless skill in a world where
logic isn't always recognized!  My usual comment when I point this out
has been, "It's a cute puzzle, but you have to take it with a grain of
salt."

I've seen several versions of the puzzle.  Here is an interesting
variation I dealt with earlier this year:

Question:

A census taker knocks on the door and a woman answers the door.  She 
informs the census taker that she lives in the house with her three 
sons.  The census taker asks for the ages of the boys and the women 
informs him that the product of their ages equals 36 and the sum of 
their ages equals the address of the house next door.  He finds this 
rather odd, but walks to the house next door, realizes that he needs 
more information, and returns.  He asks for another hint, and the 
women tells him that only her oldest son was born in a leap year. 
How old are the three sons?  Explain your answer and show all work!

My answer:

The main idea is that knowing the product to be 36 is not enough, even
WITH the knowledge of the sum of the ages.  If you list possible
products that would give 36, you can find a couple sums for which this
would be true -- and one for which the added knowledge that the oldest
son is the only one born in a certain year would clear up the uncertainty.

This version of the puzzle fixes the interesting "error" in the 
usual formulation in which you are supposed to deduce that the oldest 
simply EXISTS, and therefore is not a twin.  I have a twin brother and
a younger brother, and the former is definitely the OLDEST son; his 
mere existence as such is not enough.  The fact that he was born in 
the same year as I would do the trick.

I suppose there is still one small loophole: twins could be born in
different years, but still be the same age now.  But at least this
version shows that someone has noticed the problem!

If you have any further questions, feel free to write back.

- Doctor Peterson,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
Middle School Logic

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/