Illinois State University Mathematics Department

MAT 409: Topics in Algebra and Combinatorics for K-8 Teachers





Notes on Assignment #1







Chapter 2 Section 1 (pp 20-22)
  • #2: 9 orders required
    • The worst-case scenario is that two orders are placed for each of the one-topping pizzas. This would require 8 orders. The 9th order, however, will assure that at least one of the one-topping options is ordered at least three times.

  • #4: 70 students required
    • The worst-case scenario is that 23 students enroll in each of the three electives. This would require 69 students. The 70th student, however, will assure that at least one of the elective options is selected by at least 24 students.

  • #6: 201 letters required
    • The worst-case scenario is that 4 letters are distributed into each of the 50 mailboxes. This would require 200 letters. The 201st letter, however, will assure that at least one of the faculty has at least 5 letters in his or her mailbox.

  • #8: a = 249 apples
    • Imagine that a = 5 apples. This means that each box contains no more than 5 apples. Thus, each box contains c apples, where c is from the set {1,2,3,4,5}. Note that c cannot be 0. When a = 5, the first five boxes we look at could contain a different number of apples. The sixth box, however, forces there to be at least two boxes with an equal number of apples. Likewise, by the 11th box, we can be assured of at least three boxes containing the same number of apples, again for a = 5. This example highlights at least two things: (i) the value of a must be greater than 5, for we have 500 boxes, and (ii) we know that 2a + 1 = 11 for a = 5.

We now work backwards. We know that 2a + 1 = 500 fits the given conditions. However, there is no counting-number solution to 2a = 499, so we must solve 2a + 1 <= 500. Here, 249 apples is the largest value of a that satisfies the inequality. Put this into the context and check it. When a = 249 apples, each box contains no more than 249 apples. Thus, each box contains c apples, where c is from the set {1,2,3, . . . , 248, 249}. When a = 249, the first 249 boxes we look at could contain a different number of apples. The 250th box, however, forces there to be at least two boxes with an equal number of apples. Likewise, by the 499th box, we can be assured of at least three boxes containing the same number of apples, again for a = 249. Try a = 250 to assure yourself that 249 is the largest possible value of a to fit this situation.

  • #10: Click here for two different proofs!
    • Here, the worst-case scenario is to push the points as far from the center of this unit circle (i.e., radius r = 1 unit) as possible. Imagine a regular hexagon inscribed on a unit circle, with a radius drawn to each vertex of the hexagon. You can show that this hexagon is composed of six non-overlapping congruent equilateral triangles of side length 1 unit. Now begin distributing the six points in question. If you happen to position more than one point in or on any single equilateral triangle, those two points must be no more than one unit apart, given the dimensions of each equilateral triangle. Therefore, we must have one point in each of the six equilateral triangles.
       
  • #12: With five people in the group, labeled A, B, C, D, and E, each must have from 0 to 4 acquaintances, because you cannot be acquainted with yourself. Let us consider, then, two cases that have nothing in common.
    • Case I: Person A in the group has 0 acquaintances. If someone else in the group also has 0 acquaintances, we are done, for those two share the same number of acquaintances. Thus, for Case I, it must be that Person A is the only one with 0 acquaintances. If that is true, then each of the remaining four people must have from 1 to 3 acquaintances, for none of those four can have 0 acquaintances (previously considered) and none can have 4 acquaintances (else would be acquainted with Person A). By the Pigeonhole Principle, at least two of these four people must have the same number of acquaintances.
    • Case II: No one in the group has 0 acquaintances. If this is true, then each of the five people has from 1 to 4 acquaintances. Again by the Pigeonhole Principle, at least two of these five people must have the same number of acquaintances.
    • To generalize this, we apply the same logic to a group with n people.
    • Case I: Person A in the group has 0 acquaintances. If someone else in the group also has 0 acquaintances, we are done, for those two share the same number of acquaintances. Thus, for Case I, it must be that Person A is the only one with 0 acquaintances. If that is true, then each of the remaining n-1 people must have from 1 to n-2 acquaintances, for none of those can have 0 acquaintances (previously considered) and none can have n-1 acquaintances (else would be acquainted with Person A). By the Pigeonhole Principle, at least two of these n-1 people must have the same number of acquaintances.
    • Case II: No one in the group has 0 acquaintances. If this is true, then each of the n people has from 1 to n-1 acquaintances. Again by the Pigeonhole Principle, at least two of these n people must have the same number of acquaintances.

  • #15:
    • Five men cast in three shows each represents 15 parts in all. If we distribute those parts to each of the seven shows, we could distribute 14 parts with two parts in each of the seven shows. The 15th part, however, must now be distributed to one of the seven existing shows, thereby resulting in a show casting at least three men.

  • #16:
    • Divide each of the 12 natural numbers by 11 and consider each remainder. Possible remainders include integers 0 through 10. By the Pigeonhole Principle, at least two of the 12 natural numbers must share a remainder. Suppose these two natural numbers are N = 11j + R and M = 11k + R, where R is the common remainder, j and k are non-negative integers. Without loss of generality, suppose N > M. Then the difference D = N - M = (11j + R) - (11k + R) = 11(j - k). Because D has 11 as a factor, and j-k is a positive integer, D is divisible by 11.

  • #17: (a variation of a proof shown on this document)
    • Label the ages a1 through a33, and arrange these ages in ascending order. We have a1 + a2 + a3 + ... + a33 = 430, and we have a1 <= a2 <= a3 <= ... <= a33. Now concentrate on the ages of the 20 oldest people, and suppose that the sum of their ages is no more than 260: a14 + a15 + ... + a33 <= 260. Under this assumption, with this group of 20 people, the average age is at most 13 years. If this is true, then a14 <= 13. But if this holds, then none of a1 through a13 is greater than 13, and we can conclude that a1 + a2 + a3 + ... + a13 + a14 + ... + a33 = (a1 + a2 + a3 + ... + a13) + (a14 + ... + a33) <= 13*13 + 260 = 429 < 430, but we know the sum of all ages is 430. This contradiction stems from our assumption that the sum of the 20 oldest in the group was 260 or less. This assumption, therefore, is false, and we can conclude that the sum of the 20 oldest is more than 260. This is what we sought to show.