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 nnnn 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 = {a 1, a 2, a 3, a 4, a 5 } 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.