Math 3221 Number Theory - Homework Until Test 2 Worksheet With Answers - Philipp Braun Page 12
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1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
169 9
3. Find the last two digits of the number 9
.
9
2
4
4
4
9
Solution. 9
(9
)
9
81
9
1
9
9 (mod 10), i.e. 9
= 10k + 9 for an integer
9 9
9 9
10k+9
10k+9
10k
9
10
k
9
9
k
9
k, and therefore 9
= 9
. So 9
9
9
9
(9
)
9
(9
9)
9
(mod
9
3
3
3
100). Since 9
9
9
9
729 729 729
29 29 29
841 29
41 29
1189
89
9
k
9
k
k
k
(mod 100), (9
9)
9
(89 9)
89
(801)
89
1
89
89 (mod 100). Therefore
9 9
the last two digits of 9
are 89.
4. Without performing the divisions, determine whether the integers 176,521,221 and 149,235,678
are divisible by 9 or 11.
Solution. Using Theorem 4.5, we only have to check if the sum of the digits of the
numbers are divisible by 9 to check divisibility by 9. 1 + 7 + 6 + 5 + 2 + 1 + 2 + 2 + 1 = 27,
which is divisible by 9, so 176,521,221 is divisible by 9. 1 + 4 + 9 + 2 + 3 + 5 + 6 + 7 + 8 = 45,
which is also divisible by 9, so 149,235,678 is divisible by 9, as well.
To check divisibility by 11, we use Theorem 4.6: 1 7 + 6 5 + 2 1 + 2 2 + 1 =
3, which
is not divisible by 11, so 176,521,221 is not divisible by 11. 1 4+9 2+3 5+6 7+8 = 9,
which is also not divisible by 11, so 149,235,678 is not divisible by 11.
7. Establish the following divisibility criteria:
(a) An integer is divisible by 2 if and only if its units digit is 0, 2, 4, 6, or 8.
(b) An integer is divisible by 3 if and only if the sum of its digits is divisible by 3.
(c) An integer is divisible by 4 if and only if the number formed by its tens and units digits is
divisible by 4.
(d) An integer is divisible by 5 if and only if its units digit is 0 or 5.
Proof. (a) An integer n is divisible by 2 if and only if there is an integer k such that
n = 2k. Let l be that number out of k, k
1, k
2, k
3, and k
4 that is divisible by
5. Since 2l is divisible by 10, it is congruent 0 (mod 10). k is equal to one of the numbers
l, l + 1, l + 2, l + 3, and l + 4. In case k = l, n
2k
2l
0 (mod 10), in case k = l + 1,
n
2k
2(l + 1)
2l + 2
2 (mod 10), in case k = l + 2, n
2k
2(l + 2)
2l + 4
4
(mod 10), in case k = l + 3, n
2k
2(l + 3)
2l + 6
6 (mod 10), and finally in case
k = l + 4, n
2k
2(l + 4)
2l + 8
8 (mod 10), so the units digit of n is one of the
numbers 0, 2, 4, 6, and 8. Trivially, if n ends with 0, 2, 4, 6, or 8, it is even and therefore
divisible by 2.
(b) Let n be an integer. There are m
Z, a
, a
, . . . , a
0, 1, . . . , 9 such that n =
0
1
m
m
m
m
k
10
a
+
+ 10a
+ a
. Let P (x) =
a
x
. Then P (10) = n and P (1) =
a
, the
m
1
0
k
m
k=0
k=0
sum of the digits of n. Since 1
10 (mod 3), P (1)
P (10) (mod 3) by Theorem 4.4,
which implies the statement.
(c) Let n be an integer. There are m
Z, a
, a
, . . . , a
0, 1, . . . , 9 such that n =
0
1
m
m
10
a
+
+ 10a
+ a
. Since 100 = 4 25, 100
4 25
0 25
0 (mod 4). Therefore
m
1
0
12
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