During a month with $30$ days a baseball team plays at least a game a day; but no more than $45$ games. Show that there must be a period of some number of consecutive days during which the team must play exactly $14$ games.
- My Solution:Let the number of games played on day $i$ be $g_i$. Then, the total number of games played starting on day $m$, and ending on day $n$ is $$ g_m+g_{m+1}+\cdots+g_n . $$ Now, define s, to be the total number of games played during the first i days. That is, $$ \begin{array}{l} s_1=g_1 \\ s_2=g_1+g_2 \\ \vdots \\ s_{30}=g_1+g_2+\cdots+g_{30} \end{array} $$ Now, note that $$ \begin{aligned} g_m+g_{m+1}+\cdots+g_n &=\left(g_1+\cdots+g_n\right)-\left(g_1+\cdots g_{m-1}\right) \\ &=s_n-s_{m-1} \end{aligned} $$ Therefore, we need to show that is that $s_n-s_{m-1}$ takes on the value $14$ for some $m$ and $n$ between $1$ and $30$ . What we're given is that the total number of games played on all thirty days is no more than $45$ , which tells is that each $s_i$ is at most $45$. Furthermore, we're told that the team plays at least a game a day, which means that $s_1, s_2, \ldots, s_{30}$ are distinct integers between $1$ and $45$ . The $s_i$ are going to be the pigeons in this question. Now, let our pigeonholes be the following sets: $$ \begin{array}{l} \{1,15\},\{2,16\}, \ldots\{14,28\} \\ \{29,43\},\{30,44\},\{31,45\} \\ \{32\},\{33\} \ldots\{42\} \end{array} $$ We're picking $30$ numbers $s_1, s_2, \ldots, s_{30}$ out of the set $\{1,2, \ldots, 45\}$. Counting, we see that that we have exactly $14+3+11=28$ pigeonholes, Therefore, some two of the $s_i$ are going to be in the same pigeonhole. By our definition, a pair of numbers in a pigeonhole differ by exactly 14 - this means well have an $s_i$ and $s_j$ such that $s_j-s_i=14$. But as noted above, this mean that the total number of games played on days $i+1, i+2, \ldots, j$ is exactly $14$, as required.
If my solution seems right then , can we say that this problem is not having optimal values , that is we can have atmost $57$ games rather than $45$ since then we would be having the $30$ pigeons and $29$ holes ? Or we can have atmost $x$ games where $x$ lies between $1$ to $29$ isnt ? Not only $14$ isnt ? Atmost $30$ one not true for always and above $30$ too
I have not found a flaw in your proof. However, the following proof motivates the choice of the numbers $30, 45, 14$.
Let $x_i$ be the number of games that have been played by the $i$th day of the month. Since at least one game is played each day and a total of $45$ games are played over the course of the month, $$1 \leq x_1 < x_2 < x_3 < \ldots < x_{30} = 45$$ Define a new sequence by $y_i = x_i + 14$. Since the $x_i$ are distinct, so are the $y_i$. Moreover, $$15 \leq y_1 < y_2 < y_3 < \ldots < y_{30} = 59$$ Taking the union of the two sequences gives $60$ integers between $1$ and $59$ inclusive, so there must be some $x_i$ and $y_j$ such that $x_i = y_j$, where $j < i$. Since $x_i - y_j = 0, x_i - (x_j + 14) = 0$, so $x_i = x_j + 14$. Thus, there is a stretch of days in which the team plays exactly $14$ games, namely $x_{j + 1}, x_{j + 2}, \ldots, x_i$.