Difference between revisions of "2011 UNCO Math Contest II Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | + | The smallest number in both sets is 101. This is because it is both 1^2+100 and 10^2+1. Subtracting 10-1 gives us 9. This means that all other numbers in both sets must be less than 9 apart, because squares get further apart the bigger they get. Following a geometric sequence gives us 9,3,1 so the squares must be that far apart. This means the next term is 325, because it is 15^2+100 and 18^2+1. There is only one left, which is 2501, because it is 49^2+100 and 50^2+1. | |
== See Also == | == See Also == | ||
Revision as of 12:57, 21 December 2014
Problem
Let 
be the set of all positive squares plus 
and
be the set of all positive squares plus 
.
(a) What is the smallest number in both 
and 
?
(b) Find all numbers that are in both and .
Solution
The smallest number in both sets is 101. This is because it is both 1^2+100 and 10^2+1. Subtracting 10-1 gives us 9. This means that all other numbers in both sets must be less than 9 apart, because squares get further apart the bigger they get. Following a geometric sequence gives us 9,3,1 so the squares must be that far apart. This means the next term is 325, because it is 15^2+100 and 18^2+1. There is only one left, which is 2501, because it is 49^2+100 and 50^2+1.
See Also
| 2011 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
