How to prove that there is a string of consecutive days in which a factory produces exactly $15$ tables The problem is about "Generalized Pigeonhole Principle" from the Combinatorics textbook, and I really don't know how to solve it.
The following problem was translated from Thai into English. 

The factory wants to produce $44$ tables in $30$ days, each day must produce at least $1$ table. Prove that there must be a string of some number of consecutive days in which this factory produces exactly $15$ tables.

In my textbook, there is also a hint said 

For $i$ = $1, 2, 3, ... , 28$ , then $x$$i$ is the value of tables produced from first day to the $i$th day.

Can you help me?
 A: Hint:  Let $x_i$ be the number of tables produced through day $i$.  Since at least one table is produced each day, the $30$ numbers in the sequence $\{x_1, x_2, x_3, \ldots, x_{30}\}$ are distinct and satisfy the inequalities 
$$1 \leq x_1 < x_2 < x_3 < \ldots < x_{30} = 44$$
Let $y_i = x_i + 15$.  Since the $x_i$ are distinct, the $30$ numbers in the sequence $\{y_1, y_2, y_3, \ldots, y_{30}\}$ are distinct and satisfy the inequalities
$$16 \leq y_1 < y_2 < y_3 < \ldots < y_{30} = 59$$
The union of the two sequences contains $60$ positive integers, none of which is larger than $59$.  By the Pigeonhole Principle, there must exist $i$ and $j$ such that $x_j = y_i$.  What conclusion can you draw?
A: Consider the total number of tables produced after $n$ days, $x_n$, with $x_0=0$, and consider this value $\bmod 15$. Now we have $31$ different values $\{x_0,x_1,\ldots,x_{30}\}$ which means that, by the generalized pigeonhole principle, we must have three values which are equivalent $\bmod 15$ (we are trying to put more than $2\times 15$ pigeons into $15$ holes).
Find three such values that are equivalent $\bmod 15$, $x_a < x_b < x_c$ - they must each differ by $15$ (since otherwise $x_c-x_a>44$) so there are in fact at least two consecutive periods when exactly $15$ tables are made, on days $a{+}1$ to $b$ and on days $b{+}1$ to $c$.
