Math 3221 Number Theory - Homework Until Test 2 Worksheet With Answers - Philipp Braun Page 13
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15
16m
m
k
100 10
0 10
0 (mod 4) for any m
0; in other words 10
0 (mod 4), for
m
k = m + 2. Therefore 10
a
+
+ 10a
+ a
10a
+ a
(mod 4). That is why n
0
m
1
0
1
0
(mod 4)
10a
+ a
0 (mod 4). Note that 10a
+ a
is the number formed by the tens
1
0
1
0
and units digits of n to obtain the statement.
(d) Let n be an integer. There are m
Z, a
, a
, . . . , a
0, 1, . . . , 9 such that n =
0
1
m
m
k
k
k
10
a
+
+ 10a
+ a
. Since 10 = 2 5, 10
2
5
0 (mod 5), if k > 0. Hence
m
1
0
m
10
a
+
+ 10a
+ a
a
(mod 5). The statement is true since a
0 (mod 5) if and
m
1
0
0
0
only if a
= 0 or a
= 5.
0
0
2
8. For any integer a, show that a
a + 7 ends in one of the digits 3, 7, or 9.
Proof. By Division Algorithm, a = 10k + r, where k is an integer and r
0, 1, . . . , 9 .
2
2
2
2
2
Therefore a
a + 7 = 10
k
+ 2 10 r + r
10k
r + 7, which is congruent r
r + 7
(mod 10). Consider the following cases:
2
2
in case r = 0: a
a + 7
0
0 + 7
7 (mod 10),
2
2
in case r = 1: a
a + 7
1
1 + 7
7 (mod 10),
2
2
in case r = 2: a
a + 7
2
2 + 7
9 (mod 10),
2
2
in case r = 3: a
a + 7
3
3 + 7
13
3 (mod 10),
2
2
in case r = 4: a
a + 7
4
4 + 7
19
9 (mod 10),
2
2
in case r = 5: a
a + 7
5
5 + 7
27
7 (mod 10),
2
2
in case r = 6: a
a + 7
6
6 + 7
37
7 (mod 10),
2
2
in case r = 7: a
a + 7
7
7 + 7
49
9 (mod 10),
2
2
in case r = 8: a
a + 7
8
8 + 7
63
3 (mod 10), and finally
2
2
in case r = 9: a
a + 7
9
9 + 7
79
9 (mod 10). So we see that in each case
2
a
a + 7 is in one of the congruence classes [3], [7], [9] (mod 10), and therefore ends in
3, 7, or 9.
13
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