Name:
Solutions
Exam 1
Instructions. Answer each of the questions on your own paper, and be sure to show your work
so that partial credit can be adequately assessed. Put your name on each page of your paper.
1. [12 Points]
(a) List all of the elements in ℤ∗24 .
▶ Solution. ℤ∗24 = {1, 5, 7, 11, 13, 17, 19, 23}.
◀
(b) How many elements are in the group ℤ∗1080 ?
▶ Solution.
∣ℤ∗1080 ∣ = 𝜑(1080) = 𝜑(23 ⋅ 33 ⋅ 5 = 𝜑(23 )𝜑(33 )𝜑(5)
= (23 − 22 )(33 − 32 )(5 − 1) = 4 ⋅ 18 ⋅ 4 = 288.
◀
(c) How many elements are in the group ℤ12 × ℤ8 ?
▶ Solution. ∣ℤ12 × ℤ8 ∣ = ∣ℤ12 ∣ ⋅ ∣ℤ8 ∣ = 12 ⋅ 8 = 96.
◀
2. [16 Points]
(a) Use the Euclidean algorithm to compute 𝑑 = gcd(318, 714).
(b) Find integers 𝑠 and 𝑡 so that 𝑑 = 318 ⋅ 𝑠 + 714 ⋅ 𝑡.
▶ Solution. Use the matrix method for the Euclidean algorithm:
[
]
[
]
1 0 318
1 0 318
714−2⋅318=78
318−4⋅78=6
−→
−→
0 1 714
−2 1 78
]
[
9
−4 6
78−13⋅6=0
−→
−119 53 0
[
9 −4 6
−2 1 78
]
Hence 𝑑 = gcd(318, 714) = 6 and the first row of the last matrix implies that
𝑑 = 6 = 9 ⋅ 318 − 4 ⋅ 714.
◀
3. [14 Points]
(a) State the condition on 𝑎 that is necessary and sufficient to insure that 𝑎 has a multiplicative inverse in ℤ𝑛 .
▶ Solution. gcd(𝑎, 𝑛) = 1.
◀
(b) Compute the multiplicative inverse of 20 in ℤ263 . Express your answer in the standard
form of an integer 𝑏 in the range 0 ≤ 𝑏 < 263.
Math 4023
September 23, 2011
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Name:
Solutions
Exam 1
▶ Solution. Use the matrix method for the Euclidean algorithm to compute the greatest common divisor of 20 and 263:
]
[
]
[
]
[
1
0 20
79 −6 2
1 0 20
263−13⋅20=3
20−6⋅3=2
−→
−→
0 1 263
−13 1 3
−13 1 3
[
]
79 −6 2
3−26=1
−→
−92 7 1
Thus, 20 ⋅ (−92) + 7 ⋅ 263 = 1, so that the multiplicative inverse of 20 modulo 263 is
−92 ≡ 171 (mod 263).
◀
4. [14 Points]
(a) What power 𝑚 guarantees that the congruence
𝑎𝑚 ≡ 1 mod 21
is valid for all integers 𝑎 which are relatively prime to 21? Be sure to state which theorem
you are using for your conclusion.
▶ Solution. By Euler’s theorem, 𝑎𝜑(21) ≡ 1 (mod 21) whenever gcd(𝑎, 21) = 1. But
𝜑(21) = 𝜑(3)𝜑(7) = 2 ⋅ 6 = 12. Thus, 𝑎12 ≡ 1 (mod 21) so 𝑚 = 12.
◀
(b) Compute 599 mod 21.
▶ Solution. Since gcd(5, 21) = 1 Euler’s theorem gives (from part (a)) 512 = 𝑒𝑞𝑢𝑖𝑣1
(mod 21). Then
599 = 58⋅12+3 = (512 )8 53
≡ 18 ⋅ 53
(mod 21)
≡ 125 (mod 21)
≡ 20
(mod 21).
◀
5. [14 Points] Let 𝐺 = ℤ∗16 = {1, 3, 5, 7, 9, 11, 13, 15}, where, as usual the group operation is
multiplication modulo 16. Determine (with justification) if the following subsets of 𝐺 are
subgroups.
(a) 𝐻 = {1, 3, 9, 11}.
(b) 𝐾 = {1, 5, 9, 11}.
▶ Solution. The multiplication tables for 𝐻 and 𝐾 are:
𝐻
⋅
1 3 9 11
1 1 3 9 11
3 3 9 11 1
9 9 11 1 3
11 11 1 3 9
Math 4023
𝐾
⋅
1 5 9 11
1 1 5 9 11
5 5 9 13 7
9 9 13 1 3
11 11 7 3 9
September 23, 2011
2
Name:
Solutions
Exam 1
The table for 𝐻 shows that 𝐻 is closed under multiplication and each element has an inverse
(since 1 appears in each row and column). Thus 𝐻 is a subgroup. The table for 𝐾 shows
that 𝐾 is not closed under multiplication. In fact, 5 and 9 are in 𝐾 but 5 ⋅ 9 = 13 ∕= 𝐾. Thus,
𝐾 is not a subgroup.
◀
6. [14 Points] Answer the following questions concerning a cyclic group 𝐺 = [𝑎] with 𝑜(𝑎) = 24.
(a) What is the order ∣𝐺∣ of 𝐺.
▶ Solution. Since 𝐺 is cyclic with generator 𝑎, ∣𝐺∣ = 𝑜(𝑎) = 24.
◀
(b) What is the order of the element 𝑎20 .
▶ Solution. 𝑜(𝑎20 ) = 24/ gcd(20, 24) = 24/4 = 6.
◀
(c) How many subgroups of 𝐺 are there? Be sure to include {𝑒} and 𝐺 in your count.
▶ Solution. Since 𝐺 is cyclic of order 24, there is exactly one subgroup of 𝐺 for each
divisor of 24. There are 8 divisors of 24, namely 1, 2, 3, 4, 6, 8, 12, 24, so there are 8
subgroups of 𝐺.
◀
(d) Is the group 𝐺 × ℤ7 cyclic? Explain.
▶ Solution. 𝐺 is cyclic of order 24, and ℤ7 is cyclic of order 7, so the cartesian product
group 𝐺 × ℤ7 is cyclic since gcd(24, 7) = 1.
◀
7. [16 Points] Find the smallest positive solution to the system of simultaneous linear congruences:
𝑥 ≡ 9 mod 11
𝑥 ≡ 3 mod 9.
▶ Solution. By inspection, or using the Euclidean algorithm, we find that gcd(11, 9) = 1
and
9 ⋅ 5 − 4 ⋅ 11 = 1.
Then, the solutions to the simultaneous congruences are all of the form
𝑥 ≡ 9 ⋅ (9 ⋅ 5) + 3 ⋅ (−4 ⋅ 11) = 273
(mod 11 ⋅ 9).
Thus the solutions are of the form 𝑥 = 273 + 99𝑘 for 𝑘 ∈ ℤ. The smallest positive such
number is obtained by dividing 273 by 99 to get 273 = 2 ⋅ 99 + 75. Therefore, 75 is the
smallest positive solution to the simultaneous congruences.
◀
Math 4023
September 23, 2011
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