Challenging Algebra Age ProblemDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/