September 16, 2009
Section 2.1
2. A complete residue system modulo 17 can be taken to be
{
8
,

7
,
···
,

1
,
0
,
1
,
···
,
7
,
8
}
For each element, we can add suitable multiples of 17 to make it divisible by 3.
One possible answer is:
{
9
,

24
,

6
,
12
,

21
,

3
,
15
,

18
,
0
,
18
,

15
,
3
,
21
,

12
,
6
,
24
,

9
}
8. We check the values of
n
2
mod 10 for the complete residue system modulo 10
given by
{
4
,

3
,
···
,
0
,
1
,
···
,
4
,
5
}
. Clearly it suﬃces to check the following:
0
2
≡
0 mod 10
1
2
≡
1 mod 10
2
2
≡
4 mod 10
3
2
≡
9 mod 10
4
2
≡
6 mod 10
5
2
≡
5 mod 10
as stated.
20. Note that 42 = 2
×
3
×
7. Now by Fermat’s Little Theorem,
n
7
≡
n
mod 7
(1)
for any integer
n
. On the other hand, the congruences
n
7
≡
n
mod 3
(2)
and
n
7
≡
n
mod 2
(3)
is easily checked to be valid for any
n
(for example, to check (2), it suﬃces to
evaluate both sides on a c.r.s. modulo 3, say given by
{
1
,
0
,
1
}
). Combining
these three congruences, we obtain
n
7
≡
n
mod 42.
1