# Problem Of The Week # 291 - Dec 01, 2017

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#### Ackbach

##### Indicium Physicus
Staff member
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 POTW submission guidelines to find out how to submit your answers!

#### Ackbach

##### Indicium Physicus
Staff member
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:

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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