176
Solutions to Exercises
2.
Let the integers be
aI,
a2,
...
, a6
and consider the six integers
all
al
+
a2,
...
,
al
+ ... +
a6.
If
one
of
these
is
divisible by
6,
the result holds. Otherwise, there are only five possible remainders when each
of these is divided by 6 (namely,
1,
2,3,4
or 5). The PigeonHole Principle says that two of these must
give the same remainder; so for some
s
=f.
t,
al
+
a2
+ ... +
as
=
6q
+
r
al
+
a2
+ ... +
at
=
6q'
+
r
for the same remainder
r.
Assuming
s
>
t
(without loss of any generality), subtracting gives
at+l
+
... +
as
=
6(
q

q')
and the result holds.
3.
Let the integers be
aI,
a2,
...
,an
and consider the
n
integers
aI,
al
+
a2,
...
,
al
+ ... +
an.
If
one
of these is divisible by
n,
the result holds. Otherwise, there are only
n

1 possible remainders when
each
of
these is divided by
n
(namely,
1,2,
...
, or
n

1). The PigeonHole Principle says that two of
these must give the same remainder; so for some
s
=f.
t,
al
+
a2
+ ... +
as
al
+
a2
+ ... +
at
nq+r
nq'
+
r
for the same remainder
r.
Assuming
s
>
t
(without loss of any generality), subtracting gives
at+l
+
... +
as
=
n(q

q')
and the result holds.
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 Summer '10
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 Graph Theory, Division, Remainder, Integers, PigeonHole Principle

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