# Integral part of $\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}$ [duplicate]

I am tasked to find the integral part of $\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}$, where the number of $2018$ is $2018$.

My attempt:

I came up with an upper bound, which was $\sqrt{2018+\sqrt{2018+\sqrt{2018+...}}}$, where there were infinite number of $2018$. I let this be $a$, and then I have $a^2=2018+a$, giving me $45.4249374...$. My problem here is that I have no idea what to set as a lower bound. Can anyone give me a hint to solve this? Alternatively, another approach could also be provided if it is more efficient.

• Well, $\sqrt{2018} > 44.92$, so that's a pretty strong hint.....
– user296602
Jan 21, 2018 at 4:53
• My teacher says the answer is 45. Jan 21, 2018 at 4:54
• Why is there a vote to close? Jan 21, 2018 at 4:55

Begin by considering the fact that $45^2= 2025$. Then notice that your number is at least

$$\sqrt{2018 + \sqrt{2018}}.$$

Using the estimate that $\sqrt{2018} > 40$, which follows from $40^2 = 1600 < 2018$, your number is at least $\sqrt{2058} > \sqrt{2025} = 45$. Hence, the final answer is $45$.

• Thank you. Why is the lower bound not $\sqrt2018$ as you mentioned? Jan 21, 2018 at 4:56
• Well, $\sqrt{2018}$ is a lower bound. It's just a bad one.
– user296602
Jan 21, 2018 at 4:56
• Oh okay. Now I get it. Jan 21, 2018 at 4:57
• @XcoderX: That is a lower bound, but it is not good enough for our purposes. That gives you the answer is either $44$ or $45$. We need to work harder to get a better one and the answer has done that. Jan 21, 2018 at 4:58

For any $k$, if $k \le 2018$ then $\sqrt{2018 + \sqrt{2018 + \cdots + \sqrt{2018}}}$ with $k$ $2018$'s is a lower bound.

You just need the lower bound to be above $45$ and you win. It turns out $k = 2$ is good enough.