0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ It might help if you provide the question in the original language and someone who knows both languages can translate it better than you have. As it stands, it is not totally clear. In particular "prove that there is the consecutive days that"... how many consecutive days? Just two consecutive days, or a week? or what? "...this factory could produce 15 tables" well... of course it could produce 15 tables on a single day, but I would imagine you mean to say it must occur, as this is usually the point of questions like these $\endgroup$ – JMoravitz Sep 2 '17 at 3:17
  • $\begingroup$ Perhaps also you are thinking of the question "Prove that there must be a string of some number of consecutive days in which exactly 15 tables were produced." $\endgroup$ – JMoravitz Sep 2 '17 at 3:19
  • $\begingroup$ @JMoravitz Thank you for your comment. I've edited already $\endgroup$ – Supakorn Srisawat Sep 2 '17 at 3:21
  • 1
    $\begingroup$ In that case, consider the two sequences $a_i$, the number of tables made on day $i$, $x_i$ the number of tables made total so far by day $i$ (i.e. $x_i=a_1+a_2+\dots+a_i$*) and then consider the sequence $x_i$ modulo 15. With 30 days, what can you say about how many pairs $(i,j)$ with $i<j$ or triples $(i,j,k)$ exist with $i<j<k$ where $x_i\equiv x_j\pmod{15}$ or $x_i\equiv x_j\equiv x_k\pmod{15}$? (this is where pigeonhole is used) If $x_i\equiv x_j\pmod{15}$ what does that imply about the number $x_j-x_i$? What does that difference represent in terms of the sequence $a$? $\endgroup$ – JMoravitz Sep 2 '17 at 3:28
  • $\begingroup$ Similar questions: Jessica the Combinatorics Student and Jessica the Combinatorics Student part 2 $\endgroup$ – Joffan Sep 2 '17 at 22:17
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Is there any ceiling functions could I use? $\endgroup$ – Supakorn Srisawat Sep 2 '17 at 13:22
  • $\begingroup$ Why do you want to use a ceiling function? Just use the Pigeonhole Principle. Notice that $x_j = y_i = x_i + 15$ means that exactly $15$ tables are produced on days $i + 1$ through $j$. $\endgroup$ – N. F. Taussig Sep 2 '17 at 13:30
  • 1
    $\begingroup$ Sorry, I'm confused because some problems that use generalized pigeonhole principle also use ceiling functions like "In 100 people, there must be more than ⌈100÷12⌉ = 9 people who were born in the same month. $\endgroup$ – Supakorn Srisawat Sep 2 '17 at 13:37
  • $\begingroup$ I now understand the source of your question. Here, we just need to show that the sequences $\{x_1, x_2, x_3, \ldots, x_{30}\}$ and $\{y_1, y_2, y_3, \ldots, y_{30}\}$, where $y_i = x_i + 15$ must have a common element. $\endgroup$ – N. F. Taussig Sep 2 '17 at 13:59
0
$\begingroup$

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 $c$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for?Browse other questions tagged or ask your own question.