
Chapter 2 Section 1
(pp 2022)
 #2: 9 orders required
 The worstcase scenario is that two orders are
placed for each of the onetopping pizzas. This
would require 8 orders. The 9th order, however,
will assure that at least one of the onetopping
options is ordered at least three times.
 #4: 70 students required
 The worstcase 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 worstcase 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 countingnumber 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 worstcase 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 nonoverlapping 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 n1 people must have from
1 to n2 acquaintances, for none of those can
have 0 acquaintances (previously considered) and
none can have n1 acquaintances (else would be
acquainted with Person A). By the Pigeonhole
Principle, at least two of these n1 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 n1 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 nonnegative 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
jk is a positive integer, D is divisible by
11.
 #17: (a variation of a
proof shown on this
document)
 Label the ages a_{1}
through a_{33}, and arrange these ages
in ascending order. We have a_{1} + a_{2}
+ a_{3} + ... + a_{33} = 430,
and we have a_{1} <= a_{2}
<= a_{3} <= ... <= a_{33}.
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.

