Homework #3 Due: September 20, 2010
1. Determine if each of the following statements is true or false.
(a) 40 ≡ 13 mod 9
▶ Solution. True since 40 − 13 = 3 ⋅ 9.
▶ Solution. True since − 29 − 6 = − 35 = ( − 5) ⋅ 7.
(d) 132 ≡ 0 mod 11
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(b) − 29 ≡ 1 mod 7
▶ Solution. False since − 29 − 1 = − 30 = ( − 5) ⋅ 7 + 5, so − 29 − 1 is not divisible by 7.
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(c) − 29 ≡ 6 mod 7
◀
▶ Solution. True since 132 = 11 ⋅ 12.
2. Find all the solutions (when there are any) of the following linear congruences:
(a) 8 𝑥 ≡ 6 mod 14
◀
▶ Solution.
gcd(8 , 14) = 2 so there are 2 distinct solutions modulo 14. Divide the congruence by 2 to get 4 𝑥 ≡ 3 mod 7. Since [4] − 1
7 mod 7. Hence, 𝑥 ≡ 6 mod 14 or 𝑥 ≡ 13 mod 14.
= [2]
7
, it follows that 𝑥 ≡ 6
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(b) 66 𝑥 ≡ 100 mod 121
▶ Solution.
Since gcd(66 , 121) = 11 and 11 ∤ 100, there are no solutions to this congruence.
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(c) 21 𝑥 ≡ 14 mod 91
▶ Solution.
gcd(21 , 91) = 7 and 7 ∣ 14 so there are 91 / 7 = 13 distinct solutions mod 91. Dividing by 7 gives an equivalent congruence 3 𝑥 ≡ 2 mod 13, which has the solution 𝑥 ≡ 5 mod 13. Thus, the solutions mod 91 are 𝑥 ≡ 𝑟 mod 91, where 𝑟 ∈ { 5 , 12 , 19 , 26 , 33 , 40 , 47 , 54 , 61 , 68 , 75 , 82 , 89 } .
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(d) 21 𝑥 ≡ 14 mod 89
▶ Solution.
Since gcd(21 , 89) = 1 there is a unique solution mod 89. Applying the Euclidean algorithm gives 17 ⋅ 21 − 4 ⋅ 89 = 1, so that the inverse of 21 modulo
89 is 17. Multiplying the congruence by 17 gives 𝑥 ≡ 17 ⋅ 21 𝑥 ≡ 17 ⋅ 14 ≡ 238 ≡ 60 mod 89 .
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Math 4023 1

Homework #3 Due: September 20, 2010
3. Find the inverse of 13 modulo 35 and use it to solve the equation [13]
35 𝑥 = [9]
35 in ℤ
35
.
[
▶ Solution.
From the Euclidean algorithm, 3 ⋅ 35 − 8 ⋅ 13 = 1. Hence, [13]
− 8]
35
= [27]
35
. Multiplying the equation [13]
35 𝑥 = [9]
35 by [13] − 1
35
= [ − 8]
35
− 1
35
= gives 𝑥 = [ − 8]
35
[9]
35
= [ − 72]
35
= [ − 72 + 105]
35
= [33]
35
.
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4. Find the multiplicative inverses of the given elements (if possible).
(a) [91]
2565 in ℤ
2565
▶ Solution.
Use the Euclidean algorithm to write 451 ⋅ 91 − 16 ⋅ 2565 = 1. Hence,
[91] − 1
2565
= [451]
2565
.
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(b) [423]
6087 in ℤ
6087
▶ Solution.
Use the Euclidean algorithm to check that (423 , 6087) = 3 = 1 so there is not multiplicative inverse of 423 modulo 6087.
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5. Solve the following systems of linear congruences:
(a) 𝑥 ≡ 4 (mod 24) 𝑥 ≡ 7 (mod 11)
▶ Solution.
By inspection or using the Euclidean algorithm, find the equation
( − 5)24 + 11 ⋅ 11 = 1 .
Then the solutions to the simultaneous congruence are given by 𝑥 ≡ 4 ⋅ (11 ⋅ 11) + 7(( − 5)24) = − 356 (mod 264) .
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(b) 𝑥 ≡ 3 (mod 5) 𝑥 ≡ 4 (mod 7) 𝑥 ≡ 3 (mod 8)
▶ Solution.
By inspection or using the Euclidean algorithm, find the equation
3 ⋅ 5 − 2 ⋅ 7 = 1 .
Then the solutions to the first two congruences are given by 𝑥 ≡ 3 ⋅ ( − 2 ⋅ 7) + 4 ⋅ (3 ⋅ 5) = 18 (mod 35) .
Math 4023 2

Homework #3 Due: September 20, 2010
The the three congruences are equivalent the the pair of congruences 𝑥 ≡ 18 (mod 35) 𝑥 ≡ 3 (mod 8) .
By inspection or using the Euclidean algorithm, find the equation
3 ⋅ 35 − 13 ⋅ 8 = 1 .
Then the solutions of the simultaneous congruences are given by 𝑥 ≡ 18 ⋅ ( − 13 ⋅ 8) + 3 ⋅ (3 ⋅ 35) = − 1557 (mod 8 ⋅ 35) ≡ 123 (mod 280) .
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6. In ℤ
18 find all units and all zero divisors.
▶ Solution.
The units consist of all [ 𝑎 ]
18 consist of all [ 𝑎 ]
18 such that ( 𝑎, 18) = 1, while the zero divisors such that ( 𝑎, 18) > 1. Thus the units are and the zero divisors are
{ [1]
18
, [5]
18
, [7]
18
, [11]
18
, [13]
18
, [17]
18
}
{ [0]
18
, [2]
18
, [3]
18
, [4]
18
, [6]
18
, [8]
18
, [9]
18
, [10]
18
, [12]
18
, [14]
18
, [15]
18
, [16]
18
} .
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7. Find
(a) 6 76 mod 13,
▶ Solution.
6 12
6 76 = 6 6 ⋅ 12+4 = (6
≡
12 ) 6
1 mod 13 by Euler’s theorem, so write 76 = 6
⋅ 6 4 ≡ 6 4 ≡ (6 2 ) 2 ≡ 100 ≡ 9 mod 13.
⋅ 12 + 4 to get
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(b) 7 1001 mod 11,
▶ Solution.
By Euler, 7 10
7 1001 ≡ 7 1 = 7 mod 11.
≡ 1 mod 11. Since 1001 = 10 ⋅ 100 + 1 it follows that
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(c) 2 25 mod 21,
▶ Solution.
By Euler, 𝜑 (21) = 8 so 2 8 have 2 25 ≡ 2 1 = 2 mod 21.
≡ 1 mod 21. Since 25 = 8 ⋅ 3 + 1, we
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(d) 7 66 mod 120.
▶ Solution.
Since 𝜑 (120) = 32, Euler gives 7 32
2 ⋅ 32 + 2 it follows that 7 66 ≡ 7 2 = 49 mod 120.
≡ 1 mod 120. Since 66 =
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Math 4023 3

Homework #3 Due: September 20, 2010
8. Calculate 𝜑 (32), 𝜑 (33), 𝜑 (120), and 𝜑 (384). ( 𝜑 ( 𝑛 ) is the Euler 𝜙 function evaluated at 𝑛 .)
▶ Solution.
32 = 2 5 𝜑 (120) = 𝜑 (2 3 ) 𝜑 (3) 𝜑 so 4 𝜑 (32) = 2
(5) = (2 3 − 2 2 ) ⋅
5
2
−
⋅
2 4 = 16; 33 is prime so
4 = 32; 384 = 2 7 ⋅ 3 so 𝜑 𝜑 (33) = 33
(284) = 𝜑 (2
−
7 )
1 𝜑
− 32;
(3) =
(2 7 − 2 6 )(3 − 1) = 128.
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Math 4023 4