How many pages are in the first 5 chapters of the novel?

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    2017
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SUMMARY

The discussion centers on determining the number of pages in the first five chapters of a novel with six chapters, where the last chapter's pages sum to 2010. The mathematical approach involves the equation $$\sum_{k=1}^zk-\sum_{k=1}^yk=2010$$, leading to the conclusion that there are 90 pages in the first five chapters. The solution was derived by analyzing the divisors of 4020, with the only valid pair being (y, z) = (90, 110).

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Here is this week's POTW:

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A novel has 6 chapters. As usual, starting from the first page of the first chapter, the pages of the novel are numbered 1, 2, 3, 4, . . . . Also, each chapter begins on a new page. The last chapter is the longest and the page numbers of its pages add up to 2010. How many pages are there in the first 5 chapters ?

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to greg1313, kiwi, and castor28 for their correct solutions to this week's POTW, which was Problem 4 from the 2010 Fermat II exam put out by the University of Tennessee at Knoxville. greg1313's solution follows:

[sp]
We must have

$$\sum_{k=1}^zk-\sum_{k=1}^yk=2010$$

where $z$ is the number of pages in the book and $y$ is the number of pages in the first five chapters.

So,

$$z(z+1)-y(y+1)=(z-y)(z+y+1)=4020$$

The divisors of $4020$ are $1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60, 67, 134, 201, 268, 335, 402, 670, 804, 1005, 1340, 2010, 4020$. $z-y$ must be at least $5$ and at most $60$. Setting $z-y$ equal to these qualifying divisors, solving for $z$ in terms of $y$ and substituting for $z$ in the right-hand factor of our equation shows $(y,z)=(90,110)$ is the only possible solution, so there are $90$ pages in the first five chapters of the book.
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