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 <math>1^2+100</math> and <math>10^2+1</math>. 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 <math>15^2+100</math> and <math>18^2+1</math>. There is only one left, which is 2501, because it is <math>49^2+100</math> and <math>50^2+1</math>. | |
== See Also == | == See Also == |
Latest revision as of 20:00, 28 August 2015
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 and . 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 and . There is only one left, which is 2501, because it is and .
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 |