Math 3221 Number Theory - Homework Until Test 2 Worksheet With Answers - Philipp Braun Page 11
ADVERTISEMENT
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
163
3
in case r = 2: a
2
8 (mod 10),
3
3
in case r = 3: a
3
27
7 (mod 10),
3
3
in case r = 4: a
4
64
4 (mod 10),
3
3
in case r = 5: a
5
125
5 (mod 10),
3
3
in case r = 6: a
6
216
6 (mod 10),
3
3
in case r = 7: a
7
343
3 (mod 10),
3
3
in case r = 8: a
8
512
2 (mod 10), and finally
3
3
in case r = 9: a
9
729
9 (mod 10). So we see that each of the numbers
3
0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 occur as units digit of a
.
(c) We use again the same method to obtain
4
4
in case r = 0: a
0
0 (mod 10),
4
4
in case r = 1: a
1
1 (mod 10),
4
4
in case r = 2: a
2
16
6 (mod 10),
4
4
in case r = 3: a
3
81
1 (mod 10),
4
4
in case r = 4: a
4
256
6 (mod 10),
4
4
in case r = 5: a
5
625
5 (mod 10),
4
4
in case r = 6: a
6
1296
6 (mod 10),
4
4
in case r = 7: a
7
2401
1 (mod 10),
4
4
in case r = 8: a
8
4096
6 (mod 10), and finally
4
4
in case r = 9: a
9
6561
1 (mod 10). So the only possibilities for the units digit of
4
a
are 0, 1, 5, and 6.
(d) Let a be a triangular number. By problem 1.(a) of section 2.1, there is an integer n
n(n+1)
n +n
such that a =
=
. By Division Algorithm, n is of the form 10k + r, where k is
2
2
(10k+r) +10k+r
10 k +2 10 r+r +10k+r
an integer and r
0, 1,
, 9 . So a =
=
. Not that a
2
2
multiple of 5 ends either with 0 or with 5, so 5k
0 (mod 10) or 5k
5 (mod 10).
0 +0+10k
In case r = 0: a
0 + 5k (mod 10), so a
[0], [5]
2
1 +1+10k
in case r = 1: a
1 + 5k (mod 10), so a
[1], [6]
2
2 +2+10k
in case r = 2: a
3 + 5k (mod 10), so a
[3], [8]
2
3 +3+10k
in case r = 3: a
6 + 5k (mod 10), so a
[6], [1]
2
4 +4+10k
in case r = 4: a
10 + 5k
0 + 5k (mod 10), so a
[0], [5]
2
5 +5+10k
in case r = 5: a
15 + 5k
5 + 5k (mod 10), so a
[5], [0]
2
6 +6+10k
in case r = 6: a
21 + 5k
1 + 5k (mod 10), so a
[1], [6]
2
7 +7+10k
in case r = 7: a
28 + 5k
8 + 5k (mod 10), so a
[8], [3]
2
8 +8+10k
in case r = 8: a
36 + 5k
6 + 5k (mod 10), so a
[6], [1] , and finally
2
9 +9+10k
in case r = 9: a
45 + 5k
5 + 5k (mod 10), so a
[0], [5] . In every case a
2
is element of one of the congruence classes [0], [1], [3], [5], [6], and [8] (mod 10). This implies
the statement.
11
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education