The Pigeonhole Principle · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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More pigeons than holes forces a repeat

Mathematics · Discrete Mathematics · ages 19-20
Name ______________________   Date ____________
  1. Each integer is assigned to a category labelled by its remainder on division by 5. How many category labels are possible?

    Answer: ______________

  2. To use all files as pigeons and the folders as holes, which assignment condition is needed?

    • Some files are left outside all available folders.
    • Every file is assigned to all available folders.
    • Only folders already containing two files are counted.
    • Every file is assigned to exactly one folder.
  3. Four single-colour socks are drawn from a drawer whose socks come in 3 colours. What must happen?

    • Each colour appears at least once
    • All four differ
    • Two share a colour
  4. In any group of 13 people, two share a birth month. True or false?

    Circle one:   True   False

  5. Twenty tokens occupy 6 boxes. Five specified boxes each contain at most 2 tokens. What is the greatest lower bound for the number in the remaining box?

    Answer: ______________

  6. Seven distinct integers are chosen from 1 through 12. Which pigeonhole argument proves that two sum to 13?

    • The twelve integers form twelve holes, and seven selected numbers overfill those twelve holes.
    • The six disjoint pairs summing to 13 give six holes for the seven selected numbers.
    • The seven selected numbers form seven holes, and the twelve available integers overfill those holes.
  7. Seven people each receive one badge numbered 1 through 6. Which assumption starts a contradiction proof that two badge numbers agree?

    • Assume each badge number is used by at least one person.
    • Assume each badge number is used by at most one person.
    • Assume one badge number is used by all seven people.
    • Assume every person has a badge numbered from 1 through 6.
  8. Twenty objects in 6 boxes guarantee that some box contains exactly 4 objects. True or false?

    Circle one:   True   False

  9. Let m and n be positive integers, and let q and s be the integer quotient and remainder when m is divided by n: m = qn+s, 0 <= s < n. Which placement uses all m objects in n boxes and keeps every box at or below ceiling(m/n)?

    • Put q+1 objects in every one of the n boxes, with no boxes receiving fewer objects.
    • Put q+1 in s boxes and q-1 in each of the other n-s boxes, leaving no objects outside.
    • Put q+1 in s boxes and q in each of the other n-s boxes.
  10. Repeated values are allowed in seven selections from 1 through 12. Which selection disproves the claim that two selected entries must sum to 13?

    • 1, 12, 2, 3, 4, 5, 6
    • 2, 11, 3, 4, 5, 6, 7
    • 1, 1, 1, 1, 1, 1, 1
    • 3, 10, 4, 5, 6, 7, 8
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Answer key

For grown-ups. Fold this page away before handing over the rest.

More pigeons than holes forces a repeat W1-mt_wgfpmZqzd9-s1

  1. 5 · Remainders on division by five are 0,1,2,3,4. These give five categories.
  2. Every file is assigned to exactly one folder. · Each counted file must belong to exactly one counted folder. Omitting files or assigning them to several folders does not give this assignment.
  3. Two share a colour · Four socks outnumber the three colour categories, so some colour repeats.
  4. True · Twelve months are the holes and thirteen people overfill them.
  5. 10 · The five specified boxes hold at most ten tokens, leaving at least ten. Placing two in each specified box attains this lower bound.
  6. The six disjoint pairs summing to 13 give six holes for the seven selected numbers. · The six complementary pairs partition the available integers. Seven distinct selections force both members of one pair to be selected.
  7. Assume each badge number is used by at most one person. · Assuming no repeated number means at most one person per number. That allows at most six people, contradicting seven.
  8. False · The bound guarantees at least four, not exactly four. All twenty objects may occupy one box, with every other box empty.
  9. Put q+1 in s boxes and q in each of the other n-s boxes. · This placement uses s(q+1)+(n-s)q=qn+s=m objects. Its maximum is q when s=0 and q+1 otherwise, exactly ceiling(m/n).
  10. 1, 1, 1, 1, 1, 1, 1 · Seven copies of 1 are allowed when repetition is permitted. Any two entries sum to 2, so this selection has no pair summing to 13.
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