Integral part of $\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}$

Question

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..... — , Jan 21 '18 at 04:53

score 3 · Accepted Answer · answered Jan 21 '18 at 04:55

3

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 $45$.

answered Jan 21 '18 at 04:55

Thank you. Why is the lower bound not $\sqrt2018$ as you mentioned? – QuIcKmAtHs Jan 21 '18 at 04:56
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Well, $\sqrt{2018}$ is a lower bound. It's just a bad one. – Jan 21 '18 at 04:56
1

Oh okay. Now I get it. – QuIcKmAtHs Jan 21 '18 at 04:57
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@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. – Ross Millikan Jan 21 '18 at 04:58

score 3 · Answer 2 · answered Jan 21 '18 at 04:56

3

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.

answered Jan 21 '18 at 04:56

Caleb Stanford

45,895

Integral part of $\sqrt{2018+\sqrt{2018+\sqrt{...+2018}}}$

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