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3 age problems (Posted on 2006-03-24) Difficulty: 3 of 5
I A domestic partnership of two consenting adults has three children, Alice, Bob, and Charlie, and the difference between the adults' ages was the same as between Alice and Bob and between Bob and Charlie. The product of Alice and Bob’s ages equaled one parent's age. The product of Bob and Charlie’s ages equaled the other parent's age. The sum of the five ages amounted to ninety years. What was the age of each person?

II A woman has nine children(!), all born at regular intervals, and the sum of the squares of their ages (in years) was equal to the square of her own. What was the age of each child?

III Boadicea died one hundred and twenty-nine years after Cleopatra was born. Their combined life-span was one hundred years. Cleopatra died in 30 B.C.E. When was Boadicea born?

See The Solution Submitted by Jer    
Rating: 3.0000 (2 votes)

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re: solution | Comment 3 of 5 |
(In reply to solution by Bob Smith)

Good job, Bob.  I too tried the 2nd part using trial and error at first, and I had done the ages spaced by one year and by two years, then I had to go to a meeting before I tried a 3 year difference. :)

I wonder if someone can show how that could be solved without using trial and error...I guess it would be something like x^2 + (x+d)^2 + (x+2d)^2...etc = y^2, where x is the youngest kid, y is the mom and d is the difference in the kids' ages.

  Posted by tomarken on 2006-03-24 09:07:19
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