Challenging Algebra Age Problem

Date: 02/29/2004 at 02:02:44
From: brenda
Subject: algebra;brain teasers

A man has nine children whose ages are at an exact interval.  The sum 
of the squares of the ages of each is the square of his own age.  What 
is the age of each child and the man?

Date: 02/29/2004 at 15:20:12
From: Doctor Greenie
Subject: Re: algebra;brain teasers

Hi, Brenda --

This was an interesting problem to work on; I enjoyed the mental 
exercise it gave me.  But unless I am overlooking a clever trick, 
this is a college level problem rather than one for a 14-year-old.

Here is what I came up with....

Let

  a = number of years between children
  x = age of middle child
  y = man's age

Note that, in order to have a chance of solving this problem, we need 
to consider all of these three numbers to be whole numbers.

Then the ages of the children are

  x - 4a; x - 3a; x - 2a; x - a; x; x + a; x + 2a; x + 3a; x + 4a

The problem tells us that the sum of the squares of the children's 
ages is the square of the man's age.  So we have

  (x - 4a)^2 + (x - 3a)^2 + ... + x^2 + ... + (x + 3a)^2 + (x + 4a)^2

When we square the first term here, we get

  x^2 - 8ax + 16a^2

and when we square the last term we get

  x^2 + 8ax + 16a^2

When we add these two expressions, the "ax" terms cancel out.

The same thing happens with the other pairs; all the "ax" terms 
cancel out.  (This is why I chose to represent the ages of the 
children this way.)

So when we square the ages of all nine children and add, we get (you 
can do all the algebra to check if you want)

  9x^2 + 60a^2

And then the problem tells us that

  9x^2 + 60a^2 = y^2

When I look at this equation and consider how I might be able to find 
a solution, I see the perfect square terms "9x^2" and "y^2"; this 
leads me to try to rewrite this last equation as a difference of 
perfect squares and factor:

  y^2 - 9x^2 = 60a^2
  (y - 3x)(y + 3x) = 60a^2

Next, knowing that a is a whole number, I start looking at the cases 
where a = 1, a = 2, a = 3, ....  This means I am looking for products

  (y - 3x)(y + 3x) = 60(1^2) = 60
  (y - 3x)(y + 3x) = 60(2^2) = 240
  (y - 3x)(y + 3x) = 60(3^2) = 540
  ...

I need to find whole numbers x and y which satisfy one of these 
equations.  For example, for the case a = 1, we have the following 
possibilities:

  (y - 3x)(y + 3x) = (1)(60)
  (y - 3x)(y + 3x) = (2)(30)
  (y - 3x)(y + 3x) = (3)(20)
  (y - 3x)(y + 3x) = (4)(15)
  (y - 3x)(y + 3x) = (5)(12)
  (y - 3x)(y + 3x) = (6)(10)

The difference between the two factors on the left is 6x.  With x 
being a whole number, that means the difference between the two 
factors on the right must be a multiple of 6.  This condition is not 
satisfied for any of these factorizations.

So next we look at the case for a = 2, in which case we have

  (y - 3x)(y + 3x) = 60(2^2) = 240

Again in this case, none of the factorizations of 240 satisfies the 
requirement that the difference of the factors is a multiple of 6.

But in the case for a = 3, in which case we have

  (y - 3x)(y + 3x) = 60(3^2) = 540

we find two factorizations which satisfy that condition:

  (y - 3x)(y + 3x) = (6)(90)
  (y - 3x)(y + 3x) = (18)(30)

The second of these leads to an unacceptable solution in which the 
ages of some of the children are negative numbers.  The first of 
these, however, gives us a solution to the problem.

  y + 3x = 90
  y - 3x =  6
  -----------
      6x = 84
       x = 14

and then

       y = 48

So our solution is

  a = 3
  x = 14
  y = 48

With the definitions of these variables, this tells us our solution is

  children's ages:  2, 5, 8, 11, 14, 17, 20, 23, 26
  man's age:  48

If you do all the arithemetic, you will indeed find that the sum of 
the squares of these children's ages is equal to the square of the 
man's age.

I hope this all helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/