Question

2000 < √n+(n+2) < 2009
no. of positive integers possible?​

Answer 1

65

We know that 2^10 = 1024 < 2000 < 2048 = 2^11

Since 2^10+2^9 = 1536 < 2000, any combination of 2^m + 2^n < 2000 (as long as both m or n are not 10)

So, we would have the following possibilities;

2^10 + 2^n, (n = 1,2….9)

2^9 + 2^n, (n = 1,2….8)

And so on;

Therefore, the answer is (10*11)/2 = 65.