Math 55: Discrete Mathematics, Fall 2008
Homework 7 Solutions
5.2: 12. 26, since there are 25 possibilities for the pair (
a
mod
5
, b
mod
5).
14. (a) Consider the list of 14 integers
x
1
, . . . , x
7
,
11

x
1
, . . . ,
11

x
7
. All 14 are between
1 and 10, so there must be at least four equal pairs, since we cannot have three integers equal
on this list, and we could have at most a list of 13 with 0 to 3 equal pairs. An equal pair
must be of the form
x
i
= 11

x
j
, that is,
x
i
+
x
j
= 11. The same
x
i
and
x
j
gives a second
equal pair
x
j
= 11

x
i
, but since we have four equal pairs, we must have two solutions
x
i
+
x
j
= 11
(b) No. A counterexample is
{
1
, . . . ,
6
}
in which the only pair that sums to 11 is 5 + 6.
* 36. As noted in the hint, we cannot have a person who knows no one else and a person
who knows everyone else.
So if
k
(
x
) is the number of other people that person
x
knows,
then all the values
k
(
x
) are either in the set
{
0
,
1
, . . . , n

2
}
(if nobody knows everyone) or
in
{
1
,
2
, . . . , n

1
}
(if nobody knows no one else). Either way, there are
n
values in a set of
n

1 possibilities, so two must be equal.
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 Fall '08
 STRAIN
 Math, Natural number, Prime number, 2k, 0 K, equal pairs

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