Book divided inot Chapters

If this were an "American" textbook, I would interpret "sum of the pages" as meaning the sum of the page numbers. But in Britain, it is common to use "sum" to mean any kind of arithmetic calculation and so I agree that they mean "total number of pages.
 
I have found a similar problem on the net but I do not understand the workings out. I really want to get an understanding of what is going on...

https://sites.google.com/site/sundaytimesteasers/teaser-index-1/2013-q2/teaser-2642
First of all, they are assuming that the order of the chapters is known. If you'll remember, I asked whether that was specified in the problem. That reduces the complexity involved. (I must admit that I do not see why that assumption is justified by the statement of the problem.)

Second, this problem clearly involves the sums of the page numbers so that uncertainty is eliminated.

Third, the mention of Pell's equation means that at least one method of solution involved a Diophantine equation, as I suggested yesterday. Unfortunately, the solution of such equations is above my pay grade.

Another solution goes like this.

Let F = the number of pages in the first section.
Let S = the number of pages in the second section.
Let T = the number of pages in the third section.
Let X = F + S.

So the first section starts on page 1 and ends on page F.
The second section starts on page F + 1 and ends on page F + S.
The third section starts on page F + S + 1 and ends on page 500.

F + S + T = 500. So X + T = 500. So X = 500 - T.

We are told that the sum of the page numbers in the second section is exactly twice the sum of the page numbers in the third section.

So \(\displaystyle \dfrac{(F + S)(F + S + 1)}{2} - \dfrac{F(F + 1)}{2} = 2\left(\dfrac{500 * 501}{2} - \dfrac{(F + S)(F + S + 1)}{2}\right) \implies\)

\(\displaystyle \dfrac{X(X + 1)}{2} - \dfrac{F(F + 1)}{2} = 2\left(\dfrac{500 * 501}{2} - \dfrac{X(X + 1)}{2}\right) \implies\)

\(\displaystyle X(X + 1) - F(F + 1) = 2\{500 * 501 - X(X + 1)\} \implies\)

\(\displaystyle X(X + 1) - F(F + 1) = 501000 - 2X(X + 1)\implies\)

\(\displaystyle F^2 + F = 3X^2 + 3X - 501000 \implies\)

\(\displaystyle F^2 + F + 0.25 = 3X^2 + 3X - 500999.75 \implies\)

\(\displaystyle F + 0.5 = \sqrt{3X^2 + 3X - 500999.75} \implies\)

\(\displaystyle 2F + 1 = 2\sqrt{3X^2 + 3X - 500999.75} = \sqrt{12X^2 + 12X - 2003999} \implies\)

\(\displaystyle F = \dfrac{- 1 + \sqrt{12X^2 + 12X - 8 * 500^2 - 8 * 500 + 1}}{2}.\)

I am with him up to here. Now F must be a positive integer. And X must be a positive integer < 500. He does not say how he finds four possible values of X. I wrote a program and found them. So, at the end of the day, I had to fall back on programming.

They are F = 45, T = 91. T is more than twice F

F = 192, T = 77. T is not more than twice F.

F = 234, T = 70. T is not more than twice F.

F = 332, T = 49. T is not more than twice F.

So the answer is

F = 45
S = 364, first page 46, last page 409, sum of 46 through 409 = 82810
T = 91, first page 410, sum of 410 through 500 = 41405
 
First of all, they are assuming that the order of the chapters is known. If you'll remember, I asked whether that was specified in the problem. That reduces the complexity involved. (I must admit that I do not see why that assumption is justified by the statement of the problem.)

Second, this problem clearly involves the sums of the page numbers so that uncertainty is eliminated.

Third, the mention of Pell's equation means that at least one method of solution involved a Diophantine equation, as I suggested yesterday. Unfortunately, the solution of such equations is above my pay grade.

Another solution goes like this.

Let F = the number of pages in the first section.
Let S = the number of pages in the second section.
Let T = the number of pages in the third section.
Let X = F + S.

So the first section starts on page 1 and ends on page F.
The second section starts on page F + 1 and ends on page F + S.
The third section starts on page F + S + 1 and ends on page 500.

F + S + T = 500. So X + T = 500. So X = 500 - T.

We are told that the sum of the page numbers in the second section is exactly twice the sum of the page numbers in the third section.

So \(\displaystyle \dfrac{(F + S)(F + S + 1)}{2} - \dfrac{F(F + 1)}{2} = 2\left(\dfrac{500 * 501}{2} - \dfrac{(F + S)(F + S + 1)}{2}\right) \implies\)

\(\displaystyle \dfrac{X(X + 1)}{2} - \dfrac{F(F + 1)}{2} = 2\left(\dfrac{500 * 501}{2} - \dfrac{X(X + 1)}{2}\right) \implies\)

\(\displaystyle X(X + 1) - F(F + 1) = 2\{500 * 501 - X(X + 1)\} \implies\)

\(\displaystyle X(X + 1) - F(F + 1) = 501000 - 2X(X + 1)\implies\)

\(\displaystyle F^2 + F = 3X^2 + 3X - 501000 \implies\)

\(\displaystyle F^2 + F + 0.25 = 3X^2 + 3X - 500999.75 \implies\)

\(\displaystyle F + 0.5 = \sqrt{3X^2 + 3X - 500999.75} \implies\)

\(\displaystyle 2F + 1 = 2\sqrt{3X^2 + 3X - 500999.75} = \sqrt{12X^2 + 12X - 2003999} \implies\)

\(\displaystyle F = \dfrac{- 1 + \sqrt{12X^2 + 12X - 8 * 500^2 - 8 * 500 + 1}}{2}.\)

I am with him up to here. Now F must be a positive integer. And X must be a positive integer < 500. He does not say how he finds four possible values of X. I wrote a program and found them. So, at the end of the day, I had to fall back on programming.

They are F = 45, T = 91. T is more than twice F

F = 192, T = 77. T is not more than twice F.

F = 234, T = 70. T is not more than twice F.

F = 332, T = 49. T is not more than twice F.

So the answer is

F = 45
S = 364, first page 46, last page 409, sum of 46 through 409 = 82810
T = 91, first page 410, sum of 410 through 500 = 41405

Wow Jeff that is really smart - thanks I am going to study and digest what you have written in slower time :)
 
Wow Jeff that is really smart - thanks I am going to study and digest what you have written in slower time :)
Thanks, but hardly smart. I just worked out one of the solutions abbreviated in the link you provided.
 
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