Illinois State University Mathematics Department

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


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Notes on Assignment #1

Chapter 2 Section 1 (pp 18-20)
  • #3: 10

  • #4: 70

  • #5: There are 50 batting averages possible from .250 to .299. If every one of those 50 averages was attained by three players, that would require 150 players. By the pigeonhole princxiple, the average of the 151st player would assure that at least one of those averages was attained by at least four players. With 186 players having averages in the stated range, we will meet the condition with the 15ast player.

  • #8: The value a = 249 apples is the largest value a can take on in this situation. This tells us that among the 500 boxes, no box has more than 249 apples. If we sort the boxes according to the number of apples in each box, the first 249 boxes may all contain a different number of apples. Starting with the 250th box, we would see at least two boxes containing the same number of apples, through the 498th box that we sort. This is the worst possible scenario, that after sorting 498 boxes, we have two boxes that contain one apple, two boxes that contain two apples, and so n, up to two boxes containing 249 apples. When we sort the 499th box, however, we will be assured that at least three boxes have been identified that contain the same number of apples, by the pigeonhole principle.

  • #9: See notes from previous version of course, Supplement A problems (http://www.math.ilstu.edu/day/courses/old/305/supasoln.html)

  • #12: Same resource as for #9 above.

  • #13: Same resource as for #9 above.

  • #14: Same resource as for #9 above.

  • #15: One way to approach this problem is to consider that among the 5 actors, they appear is a total of 15 castings (i.e., three shows each) within the 7 shows of the community theatre season. If each show has no more than two men cast, that would account for at most 14 castings (7 shows, 2 men each). But the 5 men appear in 15 castings. By the pigeonhole principle, at least one of the 7 shows must have cast at least three men.

  • #16: Consider the remainder when each of the 12 natural numbers is divided by 11. Each of the 12 remainders must be in the set {0,1,2,...,10}, for we can have no remainder greater than 10 when dividing by 11. But this remainder set contains only 11 distinct values. We have remainders for 12 values. By the pigeonhole principle, at least one of these remainders must apear twice.
Suppose the duplicate remainder is r, where r is a value in the remander set, and that the two original natural numbers that resulted in this remainder are M and N, with M <= N. By the division, we have M = 11j + r and N = 11k + r, where j and k are non-negative integers and, because M <=N, j <= k. Then N - M = (11k + r) - (11j + r) = 11(k - j). But 11(k - j) is a multiple of 11, and therefore N - M, the difference of the two numbers, must be divisible by 11.
  • #17: I haven't yet determined a solution for this. Any ideas?


Syllabus
Grades & Grading
Session Notes
Assignments and Problem Sets
Tests and Quizzes