Which integer is closest to the difference?

Question

Problem: Which integer is closest to the difference? (Contest math)

$$\large\underbrace{1000\cdots}_{2020} \thinspace-\sqrt{\underbrace{99\cdots9}_{2019}\thinspace \thinspace \underbrace{00\cdots0}_{2019}}$$

My attempts:

$\underbrace{1000\cdots}_{2020}=10^{2019}$

$\sqrt{\underbrace{99\cdots9}_{2019}\thinspace \thinspace \underbrace{00\cdots0}_{2019}}=\sqrt{\underbrace{99\cdots9}_{2019}\thinspace \thinspace \underbrace{99\cdots 9}_{2019}-\underbrace{99\cdots9}_{2019}\thinspace \thinspace }=\sqrt{10^{4038}-1-10^{2019}+1}=\sqrt{10^{2019}(10^{2019}-1)}$

We have, $\underbrace{1000\cdots}_{2020} \thinspace-\sqrt{\underbrace{99\cdots9}_{2019}\thinspace \thinspace \underbrace{00\cdots0}_{2019}}=10^{2019}-\sqrt{10^{2019}(10^{2019}-1)}$

But, I can not continue from here and I don't know if my attempts are true or not.

Do my attempts correct?

Answer 1

Your attempt can be continued $$ 10^{2019} = \sqrt{10^{2019} 10^{2019}} > \sqrt{10^{2019}(10^{2019}-1)} > \sqrt{(10^{2019}-1)(10^{2019}-1)} = 10^{2019}-1 \text{,} $$ so the closest integer to $10^{2019} - \sqrt{10^{2019}(10^{2019}-1)}$ is either $0$ or $1$. So let's compare $10^{2019}(10^{2019}-1)$ with \begin{align*} (10^{2019} - 1/2)^2 &= (10^{2019})^2 - 10^{2019} + 1/4 \\ &= 10^{2019}(10^{2019}-1) + 1/4 \\ &> 10^{2019}(10^{2019}-1) \text{,} \end{align*} discovering that we are not subtracting enough for the result to be $1/2$, so the result is in the interval $(1/2,1)$ and $1$ is the closest integer to $10^{2019} - \sqrt{10^{2019}(10^{2019}-1)}$.

Answer 2

Complete the square in the radicand:

$10^{2019}-\sqrt{(10^{2019})^2-10^{2019}\color{blue}{+(1/4)}}=10^{2019}-(10^{2019}-(1/2))=(1/2)$

Infer that the closest integer to your actual defference, with the radicand being $1/4$ less, is slightly greater than $1/2$ and get the closest integer accordingly.

Answer 3

Let $x=1/10^{2019}$ and write what you've got as

$${1-\sqrt{1-x}\over x}={1\over1+\sqrt{1-x}}$$

It's easy to see that if $0\lt x\le1$ (which $1/10^{2019}$ certainly is), then

$$1={1\over1+\sqrt{1-1}}\ge{1\over1+\sqrt{1-x}}\gt{1\over1+\sqrt{1+0}}={1\over2}$$

so the nearest integer is $1$.