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/

Date: 12/13/2013 at 10:30:42
From: Ant
Subject: Census taker logic problem

Just wanted to point out that the fact that there is an oldest child does
not allow us to deduce that there cannot be two of the same age. For
example, consider two children both having the same integer age, but one
of whom is a few seconds older than the other.

If you interpret two children of the same integer age as one being older
than the other, then you must interpret their ages as continuous 
numbers -- e.g., instead of 2 years old, 2.75 years. Any positive number
may be expressed as (a + b), where a is a positive integer and b is any
positive real number smaller than one. And if we thus interpret their ages
as continuous numbers, then the product of the three children's ages will
not give an integer, since the product of any three numbers which are all
bigger than zero but smaller than one also falls into this range. The only
time we can achieve an integer (72) is if we assume that when we say "one
is older than the other" that we mean "the integer age of one is older
than the integer age other." Alternatively, it might imply that all three
children are exactly an integer number of years old at the very instant in
time.

P.S. I am not trying to give criticism, just engaging in some logical
thinking, which I enjoy.

Date: 12/13/2013 at 14:36:30
From: Doctor Peterson
Subject: Re: Census taker logic problem

Hi, Ant.

As an original contributor to this page (and a twin), I want to find out
what you are thinking to see what it adds to the thoughts I wrote. But I
confess, I'm having trouble following your reasoning.

If I understand you, you are saying that the problem is unambiguous
because if two children are the same age, there is no way to distinguish
them, so that if there is an oldest child, they can't be the same age.
That indistinguishability may be true for an outsider, but not for their
parents.

As I said on that page, there is no conflict between saying that two
children are the same age and saying that one is older. We are just
looking at the facts in two slightly different ways in making the two
statements, which is normal. No parent who says, "Here is my oldest son;
he and his brother are both 7," would be saying that they are both exactly
7; he is saying that their integer ages are both 7, AND one was born
before the other, irrespective of their "ages." No claim is made that
their birth order can be deduced from what was said about their ages.

Can you explain your thinking again?

- Doctor Peterson, The Math Forum
  

Date: 12/13/2013 at 16:29:43
From: Ant
Subject: Census taker logic problem

I am not trying to say that if a parent has 2 children both aged 7, then 
he would not know who is the eldest. Obviously, he would.

I am saying for the purposes of this question that it does not matter,
because we must interperet them as having integer ages only, which does
not allow them to be distinguished if they are both seven. I am trying to
say that the question was fine in the first place and does not need
changing.

Please do not think I am trying to argue that if you are a twin, then you
can't say that you are older than your sibling, or vice versa.

Date: 12/13/2013 at 16:47:09
From: Ant
Subject: Census taker logic problem

My point is based on the assumption that the census taker is able to
deduce their ages from the given data. If this is the case, then the
eldest must be at least 12 months older than the next oldest sibling; and
hence the question works fine as it is.

Date: 12/13/2013 at 17:49:03
From: Doctor Peterson
Subject: Re: Census taker logic problem

Hi, Ant.

Ah! Maybe that's what I've been missing in what you said.

Let's look at the problem again and see if you're right that it has a
solution even if you allow that a twin can be an oldest child.

   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?

As Dr. Tom said, with only the knowledge of the product, the possibilities
are

    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

With the additional knowledge of the sum, she still doesn't know what the
ages are, so we know that the sum must be 14, and the ages are either 2,
6, 6, or 3, 3, 8.

Now she is told that the oldest child likes chocolate pudding. Your point
is that the census taker is now sure that the oldest is not a twin because
she is told this, and therefore the oldest must be 8. My point is that she
could be WRONG in making this inference, because the father might very
well call the older of 6-year-old twins "the oldest one." So we CAN infer
that she THINKS the ages are 3, 3, 8, because this is the only reason we
can see that she would have thought she knew enough, but we CAN'T be
entirely sure that she is right! Nothing in the problem says that she
CORRECTLY deduced their ages, which is the basis of your reasoning.

So imagine if this were the puzzle's last line, instead:

   What ages did she write down?

This would be a perfectly good puzzle -- and rather ingenious; I'll have
to remember this! -- because it gives us permission to take her point of
view.

Actually, I wonder if the father, who clearly is having fun with the
census taker, might have deliberately misled her by saying something
entirely true (that the older of his 6-year-old twins likes chocolate
pudding) that seems to imply something false.

My conclusion was, "It's a cute puzzle, but you have to take it with a
grain of salt." That is, we read it within its genre (tricky puzzles) and
don't expect it to exactly fit real life. There are other issues we could
quibble about: the census taker probably needs to write down the sex of
each child, too, so she can't just walk away; and she probably would have
written down the house number before she walked in if she were really
doing her job. But we ignore that and enjoy the puzzle.

And I don't take what you're saying as criticism -- just an opportunity to
share our enjoyment of tricky puzzles.

- Doctor Peterson, The Math Forum
  
  

Date: 12/14/2013 at 03:20:22
From: Ant
Subject: Census taker logic problem

Yes, you are right: to account for my point, the question would need to
ask something like, "What did the census taker write down?"

I'd like to thank you for your response. I definitely understand the
problem much better know.

I'm not really sure how this forum works, as I found it by Googling this
problem. Do you have a page for these types of puzzles?

Thanks, Ant

Date: 12/14/2013 at 10:22:24
From: Doctor Peterson
Subject: Re: Census taker logic problem

Hi, Ant.

I don't know that this specific type is part of any collection we have,
but there are a couple parts of our site that gather together discussions
of various puzzle-type problems.

Under our FAQ (which you can get to from any page), we have a section of
Classic Problems:

  Classic Problems
    

A link at the top of the FAQ takes you to Selected Problems,

  Selected answers to common questions
     

The last section there is on Puzzles.

- Doctor Peterson, The Math Forum