2011 UNCO Math Contest II Problems/Problem 4

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Problem

Let $A = \left\{ 2,5,10,17,\cdots,n^2+1,\cdots\right\}$ be the set of all positive squares plus $1$ and $B = \left\{101, 104, 109, 116,\cdots,m^2 + 100,\cdots\right\}$ be the set of all positive squares plus $100$.

(a) What is the smallest number in both $A$ and $B$?

(b) Find all numbers that are in both $A$ and $B$.


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 (ProblemsAnswer KeyResources)
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
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