Answers to "A question of Ages"

Here is the original problem, followed by the submitted answers--enjoy the reading!!

Congratulations to the winners who answered the ages at 2, 6 and 6!!

**A QUESTION OF AGES**

**A man walks into Zeno's Coffeehouse, orders a cup of coffee, and starts chatting with
Charles. After a while, he learns that Charles has three children. "How old are your
children?" he asks. "Well," replies Charles, "the product of their
ages is 72." The man thinks for a moment and then says, "that's not enough
information." "All right," continues Charles smugly, "if you go
outside and look at the building number posted over the door to Zeno's, you'll see the sum
of the ages." The man steps outside, and after a few moments he reenters and
declares, "Still not enough!" Charles smiles and says, "My youngest plays
basketball with Maggie." How old are the children? The proud patron smiles and
promply tells Charles the ages of his children. Question: What are the
children's ages, and how did the patron deduce the ages? **

Here are the results submitted in response to Zeno's Challenge #19--they are very interesting indeed.. Comments??

Below is the result of your feedback form. It was submitted by Lauri Kultti () on Tuesday, October 27, 1998 at 04:36:06

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username: lauri.kultti@helsinki.fi

city: Helsinki

country: FINLAND

comments: All the possible ages corresponding to the product 72 (we are not dealing with decimals, I hope) are given below. The corrsponding sums are also given.

Ages Sums

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

Because the information about the product and the sum is still insufficient the only possibilities left are (2,6,6) and (3,3,8). Otherwise our man would have known the ages because all the other possibilities have unique sum - (2,6,6) and (3,3,8) both give 14.

Next information says (among other things) that there is the youngest one among the children. The (3,3,8) does not have a (unique) youngest one. So, the only possibility left is that children are 2, 6 and 6 years old.

Lauri Kultti

Department of Philosophy

University of Helsinki

Below is the result of your feedback form. It was submitted by Thomas Erlebach () on Thursday, October 29, 1998 at 04:51:33

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username: erlebach@in.tum.de

state: BY

city: Munich

url: http://www.in.tum.de/~erlebach/homepage.html

country: Germany

comments: The children's ages are 2, 6, and 6. The reasoning is as

follows. First, there are only a finite number of

possibilities for three natural numbers whose product

is 72. Most of these possibilities have a unique sum

of the three numbers; as the patron couldn't tell the

ages from knowing the sum, these possibilities are ruled

out. The remaining possibilities are 1-8-9 and 2-3-12

(both with sum 18) and 2-6-6 and 3-3-8 (both with sum 14).

Now Charles' hint that there is a "youngest" among his

children let the patron deduce the ages; therefore,

the sum must have been 14 (if the sum was 18, the

possibilities 1-8-9 and 2-3-12 would still have to

be considered). The possibility 3-3-8 is

ruled out, and the ages are 2-6-6.

Below is the result of your feedback form. It was submitted by Mathieu Weill () on Friday, October 30, 1998 at 13:23:03

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username: matweill@aol.com

state: NY

city: New York

country: USA

comments: Assuming that the children's ages are given in whole numbers and that two children with the same whole number age have the same age, the only possibility is that the children are 2, 6, and 6. 72 has 5 prime factors: 2, 2, 2, 3, and 3. So, the children's ages have to be multiples of these. This situation gives rise to 6 distinct possible threesomes: (2,2,18), (2,3,12), (2,4,9), (2,6,6), (3,4,6), and (3,3,18). Of these only two have the same sum, 14, (2,6,6) and (3,3,8) which is the only explanation for why the man was unable to give an answer knowing the sum of their ages. The last clue was that the youngest plays basketball with Maggie. Therefore, there is a youngest, and, working with the assumptions stated above, that leaves only on possible trio: 2, 6, and 6. However, the idea of an adult playing basketball with a 2 year old is not entirely logical, but not all things make sense.

Below is the result of your feedback form. It was submitted by charles krebs () on Monday, November 2, 1998 at 19:16:21

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username: ckrebs@tpolymail.calpoly.edu

state: ca

city: san luis obispo

country: USA

comments: 3,4, and 6

the product of those is 72, and the number of Zeno's is 13

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Below is the result of your feedback form. It was submitted by Josh Swartz () on Monday, November 2, 1998 at 23:24:38

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username: josh@joshswartz.com

state: Co

city: Boulder

url: http://www.joshswartz.com

country: USA

comments: I'll say the ages are 2, 6, and 6. Here's why: Knowing

the sum and the product of the ages is apparently not

sufficient to solve the problem. This means we can probably

assume the ages are one of the two factorizations of 72 that

have the same sum. These factorizationsare 2*6*6 and 3*3*8.

Now, Charles mentions that there is a youngest child. This

rules out 3, 3, and 8 and yields the answer 2, 6, and 6. The

only problem with this answer is one child can be younger

than another even if they are both "3." (e.g., one could be

37 months old and the other could be 47 months old) Also,

can a 2 year old really play basketball?

Below is the result of your feedback form. It was submitted by Ken Stange () on Thursday, November 5, 1998 at 08:01:08

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username: ken@stange.com

state: on

city: North Bay

url: ken.stange.com

country: Canada

comments: They are ages 3, 4, and 6

Presumably the youngest child that could play basketball

three years old. In x*y*z, if x=3 the only way to get the

product 72 is 3*4*6.

If the youngest is 4 or older, the product is too large.

e.g., 4 * 5 * 6 = 120

Even with twins and triplets, 4 doesn't work.

i.e., 4 * 4 * 4 = 64; 4 * 4 * 5 = 80 and of course any

other ages will be over 72 as well

Below is the result of your feedback form. It was submitted by James Holohan () on Monday, November 9, 1998 at 04:07:18

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username: j.p.holohan@student.derby.ac.uk

city: Derby

country: UK

comments: There are many ways to factorise 72, but only 8, 3, 3 and 6, 6, 2 come up with the same sum, so if we assume that the patron can work this out then the one with a "youngest" is the 6, 6, 2.

However it is perfectly feasable for three-year-olds to be born from seperate pregnancies so the question as written doesn't really hold up.

Below is the result of your feedback form. It was submitted by Christopher Horst () on Tuesday, November 10, 1998 at 15:25:49

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username: horstc@letu.edu

state: TX

city: Longview

country: USA

comments: There are several groups of three numbers whose product is 72:

1*1*72; 1*2*36; 1*3*24; 1*4*18; 1*6*12; 1*8*9;

2*2*18; 2*3*12; 2*4*9; 2*6*6; 3*3*8; 3*4*6

After finding the sum, the man could not guess only if two

groups had that sum, as in the case: 2+6+6=14; 3+3+8=14.

Of these two, only one allows for a youngest child: 2,6,6.

Therefore, these are the ages of Charles' children.

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username: Jenkins_1@bigpond.com

country: australia

comments: 9, 2, and 2.

Below is the result of your feedback form. It was submitted by Avery Johnson () on Saturday, November 14, 1998 at 13:55:50

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username: johnal01@holmes.ipfw.edu

state: In

city: Fort Wayne

country: USA

comments: I guess 3, 3, and 8, because

72 = 2 * 2 * 2 * 3 * 3

which leads to the possible combinations

2 + 2 + 18 = 22

2 + 3 + 12 = 17

2 + 4 + 9 = 15

2 + 6 + 6 = 14

3 + 3 + 8 = 14

3 + 4 + 6 = 13

Therefore the only way he could not tell their ages from

the sum would be if the building number is 14. If he can

deduce which combination is correct, it probably means he

thinks a two-year-old is too young for basketball, while a

three-year-old is not.

Below is the result of your feedback form. It was submitted by John Samaras () on Monday, November 16, 1998 at 14:33:23

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username: jmsamaras

state: GA

city: Valdosta

url: chiron.valdosta.edu/jmsamaras

country: USA

comments: The ages of Charles' children are 2, 6 and 6 (Charles has a

set of twins). Here's how the patron figured it out:

First, the product of their ages is 72. There are only six

combinations of integer ages that, when multiplied together,

produce 72. Those combinations are 2, 2, & 18; 2, 3, & 12;

2, 4, & 9; 2, 6, & 6; 3, 4, & 6; and finally 3, 3, & 8. Of

these, both 2, 6, & 6 and 3, 3, & 8 add to the same value,

namely 14. (This must be the number above Zeno's doorway,

since had it been any of the other possible sums, the patron

would have known immediately what the ages are.) Charles'

reference to the youngest in the singular eliminates the 3,

3, & 8 combination, leaving 2, 6 & 6 as the ages of Charles'

three children.

Below is the result of your feedback form. It was submitted by John Samaras () on Thursday, November 19, 1998 at 22:53:37

username: jsamaras@valdosta.edu

state: GA

city: Valdosta

url: chiron.valdosta.edu/jmsamaras

country: USA

comments: In my earlier submission, I believe that I overlooked some

other possible combinations of ages, namely 1, 1, 72 and

1, 2, 36. Since these too have unique sums, they can be

eliminated from further consideration. Apart from that,

it's hard enough to imagine Maggie playing basketball with

a two-year-old (unless one gives broad definition to what

constitutes "playing basketball"), much less doint so with

a one-year-old.