MATH 433
May 12, 2015
Final exam: Solutions
Problem 1 (15 pts.) When the number 127 · 5023 · 1812 is written out, how many zeroes are
there at the right-hand end?
Solution: 46.
The number of consecutive zeroes at the right-hand end is the exponent of the largest power of 10 that
divides our number. The prime factorisation of the given number is
127 · 5023 · 1812 = (22 · 3)7 · (2 · 52 )23 · (2 · 32 )12 = 249 · 331 · 546 .
For any integer n > 0, the prime factorisation of 10n is 2n · 5n . As follows from the Unique Factorisation
Theorem, a positive integer A divides another positive integer B if and only if the prime factorisation of A
is part of the prime factorisation of B. Hence 10n divides the given number if n ≤ 49 and n ≤ 46. The
largest number with this property is 46. Thus there are 46 zeroes at the right-hand end.
Problem 2 (20 pts.) Solve a system of congruences
x ≡ 7 mod 13,
x ≡ 8 mod 11.
Solution: x ≡ −58 mod 143 (or x ≡ 85 mod 143).
The moduli 13 and 11 are coprime. We need to represent 1 as an integral linear combination of these
moduli. It is easy to find that 6 · 11 − 5 · 13 = 1. Then one of the solutions is x = 7 · 6 · 11 − 8 · 5 · 13 = −58.
According to the Chinese Remainder Theorem, the solution set is the congruence class of −58 modulo
13 · 11 = 143.
Problem 3 (20 pts.) You receive a message that was encrypted using the RSA system with
public key (55, 27), where 55 is the base and 27 is the exponent. The encrypted message, in two
blocks, is 4/7. Find the private key and decrypt the message.
Solution: The private key is (55, 3), the decrypted message is 9/13.
First we find φ(55). The prime factorisation of 55 is 5·11, hence φ(55) = φ(5)φ(11) = (5−1)(11−1) = 40.
The private key is (55, β), where the exponent β is the inverse of 27 (the exponent from the public key)
modulo φ(55) = 40. It is easy to find by inspection that β = 3 (as 3 · 27 = 81 ≡ 1 mod 40). The standard
way to find β is to apply the Euclidean algorithm (in matrix form) to 27 and 40:
1 0 27
1 0 27
3 −2 1
→
→
.
0 1 40
−1 1 13
−1
1 13
From the first row we read off that 3 · 27 − 2 · 40 = 1, which implies that 3 is the inverse of 27 modulo 40.
Now that we know the private key, the decrypted message is b1 /b2 , where b1 ≡ 43 mod 55, b2 ≡ 73 mod 55,
and 0 ≤ b1 , b2 < 55. We find that b1 = 9, b2 = 13.
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Problem 4 (20 pts.) Find the order and the sign of a permutation π = (1 2 4)(2 3 5)(2 4 5 6).
Solution: The order is 6, the sign is −1.
First we decompose the permutation into a product of disjoint cycles: π = (1 2)(3 5 6). The order of a
product of disjoint cycles equals the least common multiple of their lengths. Therefore the order of π is 6.
Further, the transposition (1 2) is an odd permutation while the cycle (3 5 6) is an even permutation. The
product of an odd permutation and an even one is odd. Hence the permutation π is odd so that its sign is
−1.
Problem 5 (20 pts.) For any positive integer n let nZ denote the set of all integers divisible
by n. Does the set 3Z ∪ 4Z ∪ 7Z form a semigroup under addition? Does it form a group? Explain.
Solution: The set 3Z ∪ 4Z ∪ 7Z is neither a semigroup nor a group.
The set 3Z ∪ 4Z ∪ 7Z is not closed under the operation of addition. For example, the numbers 4 and 7
belong to this set while their sum 4 + 7 = 11 does not. Therefore 3Z ∪ 4Z ∪ 7Z is neither a semigroup nor a
group with respect to addition.
Problem 6 (20 pts.) Let F = {0, 1, a, b} be a field consisting of 4 elements, where 0 denotes
the additive identity element, 1 denotes the multiplicative identity element, and a, b denote the
remaining two elements. Fill in the addition and multiplication tables for the field F .
+
0
1
a
b
×
0
1
a
b
Solution:
0
1
a
b
0
1
a
b
+
0
1
a
b
0
0
1
a
b
1
1
0
b
a
a
a
b
0
1
b
b
a
1
0
×
0
1
a
b
0
0
0
0
0
1
0
1
a
b
a
0
a
b
1
b
0
b
1
a
First we fill in the multiplication table. Since 0x = 0 and 1x = x for every x ∈ F , it remains to determine
only a2 , b2 , and ab = ba. Using the fact that {1, a, b} is a multiplicative group, we obtain that ab = 1,
a2 = b, and b2 = a.
Since the number of elements in F is 4 = 22 , the field F has characteristic 2. That is, 1 + 1 = 0. Then
x + x = 1x + 1x = (1 + 1)x = 0x = 0 for all x ∈ F . Also, x + 0 = x. The rest of the addition table is filled
in using the cancellation properties, which imply that each row and each column contains all elements of F .
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Problem 7 (15 pts.) One of the groups Z20 , Z25 , and Z30 contains a subgroup isomorphic to
G5 (multiplicative group of invertible congruence classes modulo 5). Which one?
Solution: Z20 .
The group G5 consists of 4 elements. Hence any group isomorphic to G5 is also of order 4. By Lagrange’s
Theorem, the order of a finite group is divisible by the order of any subgroup. Therefore the groups Z25 and
Z30 cannot have subgroups of order 4. Thus it is the group Z20 that contains a subgroup isomorphic to G5 .
Remark. Since Z20 is a cyclic group of order 20, it has a unique cyclic subgroup of order 4
(namely, 5Z20 ). That subgroup is isomorphic to G5 since G5 is cyclic as well.
Problem 8 (20 pts.) Let f be a linear coding function defined by the generator matrix


1 0 0 1 1 0 1
0 1 0 1 1 1 0 .
0 0 1 0 1 1 1
Suppose that a message encoded by f is received with errors as 1001010 0010101 1101110. Correct
errors and decode the received message.
[(+10 pts.) if the solution relies on coset leaders and syndromes.]
Solution: 101 001 010.
The coding function f acts from B3 to B7 , where B = {0, 1}. It is given by f (w) = wG, where G is
the generator matrix, w ∈ B3 is regarded as a row vector, and the arithmetic is done modulo 2. The 8
codewords are linear combinations of rows of the generator matrix:
0000000 0010111 0101110 1001101 0111001 1011010 1100011 1110100.
The minimal weight for non-zero codewords is 4. It follows that the code can detect 3 errors and correct 1
error.
To correct the received message, we change every received word to the nearest codeword. By inspection,
each of the received words is at distance 1 from a codeword. Since the minimal distance between codewords
is 4, that is a unique closest codeword. Thus the corrected message is
1011010 0010111 0101110.
The code is systematic since the first 3 columns of the generator matrix form the identity matrix. Therefore
decoding consists of truncating the codewords to 3 digits: 101 001 010.
Alternatively, we can correct the received message using coset leaders and syndromes.
the generator matrix G into the parity-check matrix P :

1 1 0
1 1 1



0 1 1
1 0 0 1 1 0 1

A
=
G =  0 1 0 1 1 1 0  = (I3 | A) −→ P =
1 0 0
I4
0 1 0
0 0 1 0 1 1 1

0 0 1
0 0 0
3
First we transform

1
0

1

0
.
0

0
1
For any word u ∈ B7 , its syndrome is the product uP . Next we are going to build a two-column table that
contains coset leaders and their syndromes. Clearly, the zero word is a coset leader. Since the code can
correct 1 error, all words of weight 1 are coset leaders as well. These give us one half of the table (as we will
see, this half is already enough to correct our message). Seven more syndromes correspond to coset leaders
of weight 2. For the last syndrome 1011, the coset leader is of weight 3.
Coset leaders
0000000
0000001
0000010
0000100
0001000
0010000
0100000
1000000
0000011
0000101
0001001
0000110
0001010
0001100
1000010
0001011
Syndromes
0000
0001
0010
0100
1000
0111
1110
1101
0011
0101
1001
0110
1010
1100
1111
1011
Now we can start the error correction. The first received word u1 = 1001010 has syndrome 0111. The coset
leader with the matching syndrome is ũ1 = 0010000. Therefore the corrected word is u1 − ũ1 = 1011010.
Further, the second received word u2 = 0010101 has syndrome 0010. The coset leader with the matching
syndrome is ũ2 = 0000010 and the corrected word is u2 − ũ2 = 0010111. Finally, the third received word
u3 = 1101110 has syndrome 1101, the coset leader with the same syndrome is ũ3 = 1000000, and the
corrected word is u3 − ũ3 = 0101110.
Bonus Problem 9 (15 pts.) Complete the following Cayley table of a group of order 6:
∗
A
B
C
D
E
F
A B C D E F
A
D E
A
B
E
C
4
E
Solution:
∗
A
B
C
D
E
F
A
D
C
F
E
A
B
B
F
E
D
C
B
A
C
B
A
E
F
C
D
D
E
F
B
A
D
C
E
A
B
C
D
E
F
F
C
D
A
B
F
E
First we observe that E is the identity element as E 2 = E. This allows to fill in the row and the column
of E.
Next we observe that the group is not Abelian as BC 6= CB. It follows that there is no element of order 6
(such an element would generate the entire group so that the group would be cyclic and hence Abelian). By
Lagrange’s Theorem, the order of any element divides 6. Therefore it can be 1, 2, or 3. Since D 2 = A 6= E,
the element D has order 3. Then AD = DA = D 3 = E and A2 = D 4 = D.
The rest of the table can be filled in using a consequence of the cancellation property: every element of
the group appears exactly once in each row and in each column of the table.
Remark. The group in this problem is the symmetric group S(3). Namely, A stands for (1 2 3),
B stands for (1 2), C stands for (2 3), D stands for (1 3 2), E stands for id, and F stands for (1 3).
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