## A Step-by-Step Guide to Editing The 4036 Form

Below you can get an idea about how to edit and complete a 4036 Form in seconds. Get started now.

- Push the“Get Form” Button below . Here you would be transferred into a dashboard that allows you to make edits on the document.
- Pick a tool you need from the toolbar that shows up in the dashboard.
- After editing, double check and press the button Download.
- Don't hesistate to contact us via [email protected] if you need some help.

## The Most Powerful Tool to Edit and Complete The 4036 Form

## A Simple Manual to Edit 4036 Form Online

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- go to the PDF Editor Page.
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- Conduct the desired edits on your document with the toolbar on the top of the dashboard.
- Download the file once it is finalized .

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- Install CocoDoc onto your Mac device or go to the CocoDoc website with a Mac browser. Select PDF file from your Mac device. You can do so by hitting the tab Choose File, or by dropping or dragging. Edit the PDF document in the new dashboard which provides a full set of PDF tools. Save the paper by downloading.

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## PDF Editor FAQ

## What are all the ordered integer pairs (a,b) which satisfy [math]\frac{1}{a}+\frac{1}{b}=\frac{3}{2018}[/math]?

Writing the left side as a single fraction, we get [math]\frac{a+b}{ab}=\frac{3}{2018}[/math]. Since [math]\frac{3}{2018}[/math] is a fraction in its simplest form, any equivalent fraction must be of the form [math]\frac{3k}{2018k}[/math] where [math]k[/math] is an integer, but not necessarily positive. Therefore we have the simultaneous equations [math]a+b=3k[/math] and [math]ab=2018k[/math]. We can rewrite this as [math]a+\frac{2018k}{a}=3k[/math], which boils down to the quadratic equation [math]a^2-3ak+2018k=0[/math]. Using the quadratic formula, we find[math]\displaystyle a = \frac{1}{2} \left(3k \pm \sqrt{9k^2-8072k} \right) \tag*{(1)}[/math]I know you say [math](a,b)[/math] is an ordered pair, but to save time I’ll assume [math]a[/math] is the larger integer and [math]b[/math] is the smaller — at the end, we’ll just remember the pairs can be either way round. This means we pick the [math][/math][math]+[/math] sign for [math]a[/math] and [math]-[/math] sign for [math]b[/math]. Since [math]a \in \mathbb{Z}[/math], the right side of [math](1)[/math] must be an integer. Therefore [math]9k^2-8072k=x^2[/math] must be a perfect square. Multiplying through by [math]9[/math] and rearranging a little shows[math]\displaystyle (9k-4036)^2 - (3x)^2 = 16289296 \tag*{}[/math]Writing [math]16289296[/math] in its prime factorisation form and making the substitutions [math]\alpha = 9k-4036[/math] and [math]\beta = 3x[/math] shows[math]\displaystyle \alpha^2 - \beta^2 = (\alpha+\beta)(\alpha-\beta) = 2^4 \times 1009^2 \tag*{}[/math]Since each bracket is an integer, we can split into a finite number of simultaneous equations, where [math]\alpha+\beta[/math] and [math]\alpha-\beta[/math] are a factor pair of [math]16289296[/math]. To save time, we only need consider the cases when [math]\alpha+\beta[/math] is the bigger factor, and since each bracket has the same parity we can eliminate the cases when one factor is odd. I’ll walk you through one of the equations so you get the idea, but I won’t do them all here. Consider the case when [math]\alpha+\beta=8144648[/math] and [math]\alpha-\beta=2[/math]. Then [math]\alpha=4072325[/math]. Therefore [math]k=\frac{\alpha+4036}{9}=452929[/math]. Plugging this back into [math](1)[/math] shows [math]a=1358114[/math] and [math]b=673[/math]. You can use exactly the same process to solve the other equations (only some of which admit integer solutions), and make sure you remember the cases when [math]\alpha+\beta[/math] and [math]\alpha-\beta[/math] are negative.The only integer pairs are[math]\displaystyle (a,b) = (1358114,673), [/math][math][/math][math]\, (340033,674), [/math][math][/math][math]\, (2018,1009), [/math][math][/math][math]\, (672, -678048) \tag*{}[/math]Finally, don’t forget each pair can be either way round!

## What is the remainder when 2017^2018 is divided by 11?

Since [math]2017 \equiv 2^2\pmod{11}[/math] and [math]2^5 \equiv -1\pmod{11}[/math], we have[math]2017^{2018} \equiv 2^{4036} = 2 \cdot \big(2^5\big)^{807} \equiv 2 \cdot (-1) \pmod{11}[/math].The remainder is [math]9[/math]. [math]\blacksquare[/math]

## Evaluate the integral[math]\quad \displaystyle\int \dfrac{2017 x^{2016} [/math][math][/math][math]+ 2018 x^{2017}}{1 [/math][math][/math][math]+ x^{4034} [/math][math][/math][math]+ 2 x^{4035} [/math][math][/math][math]+ x^{4036}}\,dx[/math]?

We want to evaluate[math]I = \displaystyle \int \dfrac{2017 x^{2016} [/math][math][/math][math]+ 2018 x^{2017}}{1 [/math][math][/math][math]+ x^{4034} [/math][math][/math][math]+ 2 x^{4035} [/math][math][/math][math]+ x^{4036}}\,dx. \tag*{}[/math]To this end, we rewrite it as[math]I = \displaystyle \int \dfrac{2017 x^{2016} [/math][math][/math][math]+ 2018 x^{2017}}{1 [/math][math][/math][math]+ (x^{2017} [/math][math][/math][math]+ x^{2018})^2} \,dx. \tag*{}[/math]Letting [math]w = x^{2017} [/math][math][/math][math]+ x^{2018}[/math], we obtain[math]\begin{align*} I &= \displaystyle \int \dfrac{1}{1 [/math][math][/math][math]+ w^2} \,dw\\ &= \arctan{w} [/math][math][/math][math]+ C\\ &= \arctan(x^{2017} [/math][math][/math][math]+ x^{2018}) [/math][math][/math][math]+ C. \end{align*} \tag*{}[/math]

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