Download Math 170 Project #11

Math 170 Project #11
Part 2
Jeffrey Martinez
Bianca Orozco
Omar Monroy
Pigeonhole
A generalization of the pigeonhole principle
states that if n pigeons fly into m pigeonholes
and, for some positive integer k, k<n/m (the # of
pigeons doesn’t divide evenly into the
pigeonholes), then at least one pigeonhole
contains k+1 or more pigeons.
Pigeonhole Example
• Holes (m) = 4
• Pigeons (n) = 9
• Pigeons per hole (k) = 2
n
n
n
n
n
n
n
n
n
In this example, pigeonhole #1 has 3 pigeons (k+1)
9.4 Question # 37
a) Suppose a₁, a₂, … aₓ is a sequence of x integers
none of which is divisible by x. Show that at
least one of the differences aᵤ→aᵥ (for u ≠ v)
must
b) Show that every finite sequence x₁, x₂, … xₐ of n
integers has a consecutive subsequence xᵢ₊₁, xᵢ₊₂,
xᵢ₊₃, ... xj whose sum is divisible by n.
Example: n = 5: 3, 4, 17, 7, 16 has the consecutive
subsequence 17, 7, 16 whose sum is divisible by 5.
Definition of Divisibility
All numbers that are divisible by x are of the
form xn, where n is an integer.
Any integer that is not divisible by x has a
remainder c, where c is an integer greater than
zero and less than x.
xn + c, where n is an integer and c is an integer
greater than 0 and less than x.
Part a) Example
Let sequence A = {a1, a2, a3, a4, a5} of length 5. x = 5. Since none of the
elements can be divisible by 5, the elements of A are all of the form:
5n+1, 5n+2, 5n+3, or 5n+4.
Pick 4 numbers that are not divisible by 5, arbitrarily, lets pick 3, 4, 6, 7,
none of their differences are equal to 5.
3 is of the form 5(0)+3
4 is of the form 5(0)+4
6 is of the form 5(1)+1
7 is of the form 5(1)+2
Picking the 5th number, no matter what number we choose, as long as
it does not repeat a previous number of the sequence, (u ≠ v) and is
not divisible by 5, we will have a difference that is divisible by 5. For
instance let’s choose 29, which is of the form 5(5)+4, and [5(5)+4][5(0)+4] = 5(5) which is divisible by 5.
Part b) Example
Let x = 3, sequence with an element divisible by x:
{4, 6, 8}; subsequence {6} is divisible by 3
Otherwise, a sequence of numbers of which none
are divisible by 3 and have a remainder of either 1
or 2 will result in the proof from part a).
This once again is the pigeon hole principle, since
for every 3 numbers chosen that are not divisible by
3, at least 2 must have the same remainder. Since
that is true, for a consecutive subsequence of a 3
element sequence the remainders of those
numbers will at least once add up to 3.