Find the number of positive integer n, such that √n + √n+1 < 11

Question

This question is prmo 2017. Based on number theory.

Answer 1

√n + √(n+1) < 11 Subtract √(n+1) from both sides: √n < 11 - √(n+1) Square both sides: n < 121 - 22√(n+1) + n + 1 Subtract n from both sides and simplify the right hand side: 0 < 122 - 22√(n+1) Add 22√(n+1) to both sides: 22√(n+1) < 122 Divide both sides by 22: √(n+1) < 122/22 √(n+1) < 61/11 Square both sides: n+1 < 3721/121 = 30.75 Subtract 1 from both sides: n < 29.75 Therefore, positive integers 1-29 should work! We can check this: √29 + √30 = 10.86 If we go to n = 30, it should be greater than 11: √30 + √31 = 11.05 Therefore, your answer is x < 30. Hope this helps!