Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 1"

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== Problem ==
 
== Problem ==
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The Duchess had a child on May 1st every two years until she had five children. This year the
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youngest is <math>1</math> and the ages of the children are <math>1, 3, 5, 7</math>, and <math>9</math>. Alice notices that the sum of the ages
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is a perfect square: <math>1 + 3 + 5 + 7 + 9 = 25</math>. How old will the youngest be the next time the sum of
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the ages of the five children is a perfect square, and what is that perfect square?
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== Solution ==
 
== Solution ==
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We know that the sum of the ages is 25. To get another perfect square, we need to multiply it by a perfect square, so we get 100 for the next perfect square. We know their ages are (N-4)+(N-2)+(N)+(N+2)+(N+4)=100. Solving for N gives us 20, so the youngest is 20-4=16.
  
 
== See also ==
 
== See also ==
{{UNC Math Contest box|year=2014|n=II|before=First Question|num-a=2}}
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{{UNCO Math Contest box|year=2014|n=II|before=First Question|num-a=2}}
  
 
[[Category:Introductory Number Theory Problems]]
 
[[Category:Introductory Number Theory Problems]]

Latest revision as of 15:31, 23 December 2014

Problem

The Duchess had a child on May 1st every two years until she had five children. This year the youngest is $1$ and the ages of the children are $1, 3, 5, 7$, and $9$. Alice notices that the sum of the ages is a perfect square: $1 + 3 + 5 + 7 + 9 = 25$. How old will the youngest be the next time the sum of the ages of the five children is a perfect square, and what is that perfect square?


Solution

We know that the sum of the ages is 25. To get another perfect square, we need to multiply it by a perfect square, so we get 100 for the next perfect square. We know their ages are (N-4)+(N-2)+(N)+(N+2)+(N+4)=100. Solving for N gives us 20, so the youngest is 20-4=16.

See also

2014 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions