2011 UNCO Math Contest II Problems/Problem 4

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 $50^2+1$ .

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

Page

Toolbox

Search

2011 UNCO Math Contest II Problems/Problem 4

Problem

Solution

See Also