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MATH 135
Fall 2008
Assignment #7 Solutions
HandIn Problems
1. Considering
a
modulo 11, we obtain
a
≡
10
n
r
n
+ 10
n

1
r
n

1
+
···
+ 10
r
1
+
r
0
(mod 11)
≡
(

1)
n
r
n
+ (

1)
n

1
r
n

1
+
···
+ (

1)
r
1
+
r
0
(mod 11)
(since 10
≡ 
1
(mod 11))
≡
r
0

r
1
+
···
+ (

1)
n

1
+ (

1)
n
(mod 11)
Therefore,
a
≡
r
0

r
1
+
···
+ (

1)
n

1
+ (

1)
n
(mod 11), and so 11

a
if and only if
a
≡
0 (mod 11) if and only if
r
0

r
1
+
···
+ (

1)
n

1
+ (

1)
n
≡
0 (mod 11) if and only if
11

r
0

r
1
+
···
+ (

1)
n

1
+ (

1)
n
, as required.
2. For 23
b
2
a
4 to be divisible by 72, it must be divisible by 8 and by 9.
By (a), 23
b
2
a
4 is divisible by 8 if and only if 2
a
4 is divisible by 8.
Checking the possibilities, 2
a
4 is divisible by 8 only when
a
= 2 or
a
= 6.
Thus, 23
b
2
a
4 = 23
b
224 or 23
b
264.
For 23
b
224 to be divisible by 9, we need 2 + 3 +
b
+ 2 + 2 + 4
≡
0 (mod 9) or
b
≡ 
13 (mod 9)
or
b
= 5 since
b
is a digit.
For 23
b
264 to be divisible by 9, we need 2 + 3 +
b
+ 2 + 6 + 4
≡
0 (mod 9) or
b
=

17 (mod 9)
or
b
= 1 since
b
is a digit.
Thus, for 23
b
2
a
4 to be divisible by 72, we must have (
a, b
) = (2
,
5) or (6
,
1).
(We should check that each of these possibilities is indeed divisible by 72.
Note that 235 224
÷
72 = 3267 and 231 264
÷
72 = 3212 so both are indeed divisible by 72.)
3. (a) We convert to the linear Diophantine equation 1713
x
+ 2000
y
= 851 and solve using the
EEA:
0
1
2000
1
0
1713

1
1
287
1
6

5
278
5

7
6
9
1
216

185
8
30

223
191
1
1
2000

1713
0
8
Therefore, gcd (1713
,
2000) = 1, so there is a unique solution mod 2000.
Since 1713(

223) + 2000(191) = 1, then, multiplying by 851, we obtain
1713(

189773) + 2000(162541) = 851, so the solution to the congruence is
x
≡ 
189773
≡
227 (mod 2000).
(b) Since 2

1426 and 2

2000, then 2

gcd(1426
,
2000).
(In fact, gcd(1426
,
2000) = 2.)
Since 2
6 
851, then there is no solution.
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View Full Document (c) We make a table modulo 12 to solve
x
2

2
x
≡
0 (mod 12):
Modulo 12
x
≡
0
1
2
3
4
5
6
7
8
9
10
11
2
x
≡
0
2
4
6
8
10
0
2
4
6
8
10
x
2
≡
0
1
4
9
4
1
0
1
4
9
4
1
x
2

2
x
≡
0
11
0
3
8
3
0
11
0
3
8
3
so
x
≡
0
,
2
,
6
,
8 (mod 12).
(d)
Solution 1
Since 8
x
≡
12 (mod 52) and each of 8, 12 and 52 is divisible by 4, then 2
x
≡
3 (mod 13).
This also means that 2
x
≡
16 (mod 13).
Since gcd(2
,
13) = 1, we can divide out the common factor of 2 to obtain
x
≡
8 (mod 13).
Solution 2
Since gcd (18
,
52) = 4 and 4

12, then there are 4 solutions modulo 52 (or 1 solution
modulo 13).
By inspection,
x
= 8 is a solution, so the complete solution is
x
≡
8
,
8 +
52
4
,
8 + 2
(
52
4
)
,
8 + 3
(
52
4
)
(mod 52)
≡
8
,
21
,
34
,
47
(mod 52)
or
x
≡
8
(mod 13)
.
4. (a) We want to solve [23][
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This note was uploaded on 02/23/2009 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Algebra

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