Pigeonhole Principle

Problem 1: Squares

• squares

Problem 2: Handshake

• $n$ people: attend party
• some: shake hands
• let $d(i)$: no. of handshake person $i$ has made
• show: ${\sum }_{i=1}^{n}d(i)=2k$ (even)
  • trivial as there always exist
• show: there are two people $i\ne j,d(i)=d(j)$
  • $d(i)\in [0,n-1]$
    • $n$ different holes, for $n$ people
  • however, we can
  • case 1: $\exists i,d(i)=0$
    • i.e. for every $j,d(j)\in [0,n-2]$
    • as one person didn't shook hands with anyone
    • with $n-1$ people left: there exists at least one hole with 2 or more people
      • as there is a person in hole $0$ in the first place
  • case 2: $\nexists i,d(i)=0$
    • i.e. for every $j,d(j)\in [1,n-1]$
    • $n-1$ hole for $n$ people: must be double count!

Problem 3

• show that: $\exists x,y\in \mathbb{N},2024|9^x-9^y$
  • let $f:g^x\mapsto 9^x\mathrm{mod}2024$
  • $P=\{9^0,9^1,9^2,\dots \}$
    • infinite set!
      • actually: you can have $2025$ of them
      • but, as it's infinite set, there are infinite such solutions
    • $H=\{0,1,2,\dots ,2023\}$
  • $9^x\equiv 9^y\mathrm{mod}2024$
    • there must be at least 1 collision in $n+1=2025$ numbers
    • then: $9^x-9^y\equiv 0\mathrm{mod}2024$
  • 👨‍🏫 size of $9^x,9^y$: doesn't matter

Problem 4

• show that
  • $3\times 3$ grid, $\{-1,0,1\}$
  • goal: to make every row, column, and diagonal: has different sum
    • is this doable?
  • there are $7$ possible sum: $[-3,3]$
  • but there are 8 numbers: $3+3+2$ for row, column, and diagonal
    • thus, there must be at least one collision

Problem 5

• for a set of $n$ integers $X$
  • $\exists Y\subseteq X,Y\ne \varnothing \wedge n|{\sum }_{a\in Y}a$
  • attempt 1
    • $H=\{0,1,\dots n-1\}$
    • $P=$ subsets of $X$
    • $f:Y\mapsto ({\sum }_{a\in Y}a)\%n$
    • $|P|$: $2^n-1$ (non-empty)
    • $|H|$: $n$
    • while it shows: there exists a collision
      • doesn't show that
  • also, if $A\subset B$
    • and $f(A)=f(B)$
    • then $B-A$: solution to this problem
  • attempt 2
    • $X=\{x_1,x_2,\dots ,x_n\}$
    • enumerating: $\varphi ,\{x_1\},\{x_1,x_2\},,\{x_1,x_2,x_3\},\dots X$
    • i.e. $P^{\prime }=X_i:\{x_j|j\le i\}$
      • $|P^{\prime }|=n+1$
      • when $|H|=n$
    • there are two subsets $X_i<X_j;j>i$
      • s.t. ${\sum }_{x\in X_i}x\equiv {\sum }_{x\in X_j}x\mathrm{mod}n$
      • then, $({\sum }_{a\in \{x_k|i<k\le j\}}a)\%n=0$
    • however: works only if empty set is included
  • not considering an empty set:

Problem 6

• problem
  • let $X\subseteq [1,99]$; $|X|=10$
  • show that there are two disjoint non-empty proper subsets of $X$
  • w/ the same sum
  • rules
    • $Y,Z\subseteq X$
    • $Y\cap Z=\varnothing$
    • $Y=\varnothing ;Z=\varnothing$
    • $Y\ne X;Z\ne X$
  • how many proper subsets: $2^{1}0-2=1022$
    • i.e.: we have 1022 pigeons
    • $f:Y\mapsto {\sum }_{y\in Y}y$
    • $1\le {\sum }_{y\in Y}y\le 91+92+\cdots 99=855$
  • $|H|=\{1,2,\dots 855\}$
  • and there must be at least 2 non-empty proper subsets w/ same sum
    • as $|H|<|P|$
• disjoint part:
  • if ${\sum }_{x\in X}x={\sum }_{y\in Y}y$
    • then ${\sum }_{x\in X{Y}^c}x={\sum }_{y\in X^{c}Y}y$
    • as you can subtract the same number
  • Y=A</span>B
  • Z=B</span>A

Problem 7

• 15 people w/ 100 coins in total;
  • show: two of them have the same number of coins
  • if each person has different no. of coins: $\sum [0,14]$
    • thus min: $\frac{14\cdot 15}{2}=105>100$
  • two: forced to have the same no. of coins

Problem 8

• football team w/ 20 games
  • scores in all of them
  • team: scores $30$ goals in total
  • show: there is a sequence of conseq. games s.t. the team has scored:
  • exactly 9 goals
  • 👨‍🏫 problem w/ bunch of numbers: either induction / pigeon hole
  • $g_i$: no. of goals in $i$-th game
  • $1\le g_1,g_2,\dots ,g_{20}$
  • $g_1+g_2+\dots g_{20}=30$
  • $a_i={\sum }_{j\le i}{g}_j$; $a_0=0$
    • 30 different values: $1\le a_i\le 30$
  • prove: $\exists i,j$ s.t. $a_j-a_i=9$
    • and all $a_i$: distinct as
      • $g_j>0$ for all
  • let's try to avoid having $9$ difference
    • intuition: for each choice, you lose at least 1 choice
      • e.g. if you choose $a_i=15$, then you can't choose $6,24$
    • alternatively: you can "pair" numbers up s.t.
      • you can choose at most one number from them
      • e.g. $\{1,10\},\{2,11\},\dots ,\{9,18\}$ and $\{19,28\},\{20,28\}\{21,30\}$
  • 👨‍🏫 is there a more elegant way?
    • let $P\ni a_1$, another $a_1+9$
      • $P\ni a_2$, another $a_2+9$
      • $P\ni a_3$, another $a_3+9$
      • ...
      • $P\ni a_{20}$, another $a_{20}+9$
    • there are 40 numbers on left & right combined
      • where we have 30 possibilities
      • however: we must choose those two numbers from different sides
      • $a_j=a_i+9$

Problem 9

• let $X\subseteq \{1,2,\dots ,100\}$ and $|X|=16$
  • prove that there exists four different no. s.t. $a,b,c,d\in X$
    • s.t. $a+b=c+d$
    • alternatively: $a-c=d-b$
    • 👨‍🏫 preferable as subtraction has less variety than addition
  • $|X|=16$, so there are $\left(\genfrac{}{}{0ex}{}{16}{2}\right)=120$ pairs
    • if $a_1>a_2$
    • we have: $1\le a_1-a_2\le 99$
  • so there exists at least two pairs: $(a_1,a_2),(a_3,a_4)$
    • s.t. their difference: same
    • as $99<120$
  • yet, we want for distinct numbers
    • while $a_1=a_4$ can be true
  • then, we can eliminate pairs with duplicated value
    • if $a=(b+c)/2$, then we can eliminate either
      • $(a,b)$ or $(a,c)$
    • but if $a=(b^{\prime }+c^{\prime })/2$ at the same time
      • $b-b^{\prime }=c^{\prime }-c$
      • thus solution discovered

Problem 10

• 6 candidates for president
  • every pair of candidates: either friends or enemies
  • show, there are 3 candidates friend to each other
    • or enemy with each other
  • 👨‍🏫 can be represented as a graph, with edge being different color
  • statement claims: there exists at least 1 all-blue or all-red triangle
  • choose 1 candidate

Problem 10+

• IMO 1964
  • 17 people correspond: by mail with each other: one with all the rest
    • only three topics are discussed
    • each pair of correspondents: deal with 1 of these topics
    • prove: that there are at least tree people, writing to each other, about the same topic
  • choose 1 candidate
    • among mails, at least $\lceil 16/3\rceil =6$ should have the same topic
    • a
  • what if: four different..?
• also: HW