We are trying to solve a math contest problem involving page numbers and a book, which goes:
A book contains 30 stories, each starting on a new page. The lengths of the stories are 1, 2, 3, ..., 30 pages in some order. The first story starts on the first page. What is the largest number of stories that can start on an odd-numbered page?
Solving this, The first story starts and ends on page 1, while the 2nd story starts on page 2 and ends on page 3, and the 3rd story starts on page 4 and ends on page 7, and so on. After the first story, the pattern is 2 stories starting on an even page, followed by 2 stories starting on an odd-numbered page, and then 2 even numbered starting pages, and then 2 odd, and this pattern repeats until the end.
This would give us a maximum of 15 stories starting on an odd-numbered page given that there are 30 stories all in all.
However, 2 separate answer keys give the answer at 23. We are at a loss as to why this is the answer and would appreciate any help. Thanks!
Given that there are $30$ stories, Let the number of pages for $$1\text{st Story be } a_1$$ $$2\text{nd Story be } a_2$$ $$3\text{rd Story be } a_3$$ $$4\text{th Story be } a_4$$ $$5\text{th Story be } a_5$$ $$\vdots$$ $$\vdots$$ $$29\text{th Story be } a_{29}$$ $$30\text{th Story be } a_{30}$$
Note that $1$st Story starts from page $1$, because $1$st Story has $a_1$ pages, $2$nd Story will start from page numbered $1 + a_1$
Further Note that $2$nd Story starts from page $1+a_1$, because $2$nd Story has $a_2$ pages, $3$nd Story will start from page numbered $1 + a_1 + a_2$
In General Note that $\left(n-1\right)$th Story starts from page $1+a_1+a_2+a_3 \dots \dots +a_{n-2}$, because $\left(n-1\right)$th Story has $a_{n-1}$ pages, $n$th story will start from page numbered $1 + a_1 + a_2+a_3 \dots \dots +a_{n-2}+a_{n-1}$
Now all we want is some order of $1,2,3 \dots 30$ assigned to $a_1,a_2,a_3. \dots a_{30}$ such that, out of the $30$ numbers below, Most of them are odd $$1$$ $$1+a_1$$ $$1+a_1 + a_2$$ $$1 + a_1 + a_2 + a_3$$ $$\vdots$$ $$\vdots$$ $$1 + a_1 + a_2 + a_3 + \dots \dots a_{28} + a_{29}$$
This is equivalent to finding the order of $1,2,3 \dots 30$ assigned to $a_1,a_2,a_3. \dots a_{30}$ such that Maximum Numbers below are even among $$S_1=a_1$$ $$S_2=a_1 + a_2$$ $$S_3=a_1 + a_2 + a_3$$ $$\vdots$$ $$\vdots$$ $$S_{29}=a_1 + a_2 + a_3 + \dots \dots a_{28} + a_{29}$$
Note that Odd$+$Odd$=$Even and Even$+$Even$=$Even
If $S_1,S_2,S_3,S_4,S_5 \dots \dots S_15, S_{17},S_{19},S_{21} \dots S_{29} are Even$ by using following arrangement of $1,2,3 \dots 30$ assigned to $a_1,a_2,a_3. \dots a_{30}$ such that $a_1=E,a_2=E \dots \dots a_{15}=E,a_{16}=O,a_{17}=O \dots a_{29}=O$ Where $E$ denotes Even Number and $O$ denotes Odd Number. This gives $S_1,S_2,S_3,S_4,S_5 \dots \dots S_{14},S_{15}, S_{17},S_{19},S_{21} \dots S_{29}$ all even, These are $22$ in Number and 1st Chapter also starts with 1, Hence 23 Possibilites