Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 1"
m (moved 2014 UNC Math Contest II Problems/Problem 1 to 2014 UNCO Math Contest II Problems/Problem 1: disambiguation of University of Northern Colorado with University of North Carolina) |
Pi3point14 (talk | contribs) (→Solution) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 8: | Line 8: | ||
== Solution == | == 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 == | == See also == | ||
| − | {{ | + | {{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 
and the ages of the children are 
, and 
. Alice notices that the sum of the ages
is a perfect square: 
. 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 (Problems • Answer Key • Resources) | ||
| 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 | ||
