MATH 433100
June 11, 2010
Exam 1: Solutions
Problem 1 (15 pts.) Find the smallest positive integer a such that the equation
76x + 96y = a has an integer solution.
Solution: a = 4.
The sought number is the greatest common divisor of 76 and 96. Since 76 = 2
MATH 433100
June 2, 2010
Quiz 1: Solutions
Problem 1. Find all integers 1 x 100 such that gcd(x, 15) = 5 and gcd(x, 16) = 4.
Solution: x = 20 or 100.
Since x is divisible by the numbers 4 and 5, which are coprime, it is also divisible by 4 5 = 20. Hence
x
MATH 433100
June 3, 2010
Quiz 2: Solutions
1
1
11
Problem 1. Using the induction principle, prove that + + + n = 1 n for every
24
2
2
positive integer n.
First consider the case n = 1. In this case the formula reduces to
Now assume that the formula holds
MATH 433100
June 4, 2010
Quiz 3: Solutions
Problem 1. Let a and b be positive integers. Explain why gcd(a2 , b2 ) cannot be equal to 3.
Can gcd(a2 , b2 ) be equal to 8?
Solution: gcd(a2 , b2 ) cannot be equal to 3 or 8.
Let p1 p2 . . . pk be the prime fac
MATH 433100
June 8, 2010
Quiz 4: Solutions
Problem 1. Find an integer x such that 4x 18 mod 11.
Solution: x = 10 (as well as any x 10 mod 11).
To solve this linear congruence, we need to nd the inverse of 4 modulo 11. For this, we need to represent
1 as a
MATH 433100
June 9, 2010
Quiz 5: Solutions
Problem 1. Which of the values (17) and (48) of Eulers totient function is greater?
Solution: (17) = (48) = 16.
The number 17 is prime, therefore (17) = 17 1 = 16. The prime decomposition of 48 is 24 3. It
follow
MATH 433100
June 15, 2010
Quiz 6: Solutions
Problem 1. An acceptor automaton M has the set of states S = cfw_0, 1, 2, 3, 4, where 0 is the
initial state. The automaton has two acceptance states: 3 and 4. The alphabet is A = cfw_a, b and
the state transiti
MATH 433100
June 16, 2010
Quiz 7: Solutions
12345
,=
32514
for the permutations , , and = .
Problem 1.
Let =
12345
. Find the cycle decomposition
21453
Solution: = (1 3 5 4), = (1 2)(3 4 5), = (1 2 3).
The two-row representation of the permutation is obta
MATH 433100
June 17, 2010
Quiz 8: Solutions
Problem 1. Find the order of the permutation = (1 2)(2 3 5)(3 4 5 6).
Solution: has order 4.
The permutation is dened as a product of cycles that are not disjoint. To nd the order of , we
need to represent as a
MATH 433100
June 18, 2010
Quiz 9
Problem 1. Express the permutation = (1 2 4)(2 3 4 5) as a product of transpositions and
nd the sign of .
Solution: = (1 2)(2 4)(2 3)(3 4)(4 5), sgn( ) = 1.
We have (1 2 4) = (1 2)(2 4) and (2 3 4 5) = (2 3)(3 4)(4 5), hen
MATH 433100
June 22, 2010
Quiz 10: Solutions
nk
, where n and k are
0n
integers. Under the operation of matrix multiplication, does this set form a semigroup? Does M
form a group? Is the operation commutative on M ? Explain.
Problem 1.
Let M be the set of
MATH 433100
June 23, 2010
Quiz 11: Solutions
n0
, where n and k are
kn
integers. Under the operations of matrix addition and multiplication, does this set form a ring? Does
M form a eld? Explain.
Problem 1.
Let M be the set of all 2 2 matrices of the form
MATH 433100
June 24, 2010
Quiz 12: Solutions
Problem 1. Consider the multiplicative group GL(2, Z2 ) of 2 2 matrices with entries from
the eld Z2 and nonzero determinant. How many elements are there in this group?
Solution: The group GL(2, Z2 ) consists o
MATH 433100
June 29, 2010
Quiz 14: Solutions
Problem 1. List all subgroups of the group (G7 , ).
Solution: cfw_[1], cfw_[1], [6], cfw_[1], [2], [4], and cfw_[1], [2], [3], [4], [5], [6].
G7 is the multiplicative group of invertible congruence classes modu
MATH 433100
July 2, 2010
Exam 2
Problem 1 (10 pts.)
product of disjoint cycles.
Write the permutation = (1 3 5)(2 4)(1 2 3 4 5 6) as a
Solution: = (1 4)(2 5 6 3).
Keeping in mind that the composition is evaluated from the right to the left, we nd that
(1
MATH 433100
June 11, 2010
Exam 1: Solutions
Problem 1 (15 pts.) Find the smallest positive integer a such that the equation
76x + 96y = a has an integer solution.
Solution: a = 4.
The sought number is the greatest common divisor of 76 and 96. Since 76 = 2
MATH 433100
July 2, 2010
Exam 2
Problem 1 (10 pts.)
product of disjoint cycles.
Write the permutation = (1 3 5)(2 4)(1 2 3 4 5 6) as a
Solution: = (1 4)(2 5 6 3).
Keeping in mind that the composition is evaluated from the right to the left, we nd that
(1
MATH 433100
June 2, 2010
Quiz 1: Solutions
Problem 1. Find all integers 1 x 100 such that gcd(x, 15) = 5 and gcd(x, 16) = 4.
Solution: x = 20 or 100.
Since x is divisible by the numbers 4 and 5, which are coprime, it is also divisible by 4 5 = 20. Hence
x
MATH 433100
June 3, 2010
Quiz 2: Solutions
1
1
11
Problem 1. Using the induction principle, prove that + + + n = 1 n for every
24
2
2
positive integer n.
First consider the case n = 1. In this case the formula reduces to
Now assume that the formula holds
MATH 433100
June 4, 2010
Quiz 3: Solutions
Problem 1. Let a and b be positive integers. Explain why gcd(a2 , b2 ) cannot be equal to 3.
Can gcd(a2 , b2 ) be equal to 8?
Solution: gcd(a2 , b2 ) cannot be equal to 3 or 8.
Let p1 p2 . . . pk be the prime fac
MATH 433100
June 8, 2010
Quiz 4: Solutions
Problem 1. Find an integer x such that 4x 18 mod 11.
Solution: x = 10 (as well as any x 10 mod 11).
To solve this linear congruence, we need to nd the inverse of 4 modulo 11. For this, we need to represent
1 as a
MATH 433100
June 9, 2010
Quiz 5: Solutions
Problem 1. Which of the values (17) and (48) of Eulers totient function is greater?
Solution: (17) = (48) = 16.
The number 17 is prime, therefore (17) = 17 1 = 16. The prime decomposition of 48 is 24 3. It
follow
MATH 433100
June 15, 2010
Quiz 6: Solutions
Problem 1. An acceptor automaton M has the set of states S = cfw_0, 1, 2, 3, 4, where 0 is the
initial state. The automaton has two acceptance states: 3 and 4. The alphabet is A = cfw_a, b and
the state transiti
MATH 433100
June 16, 2010
Quiz 7: Solutions
12345
,=
32514
for the permutations , , and = .
Problem 1.
Let =
12345
. Find the cycle decomposition
21453
Solution: = (1 3 5 4), = (1 2)(3 4 5), = (1 2 3).
The two-row representation of the permutation is obta
MATH 433100
June 17, 2010
Quiz 8: Solutions
Problem 1. Find the order of the permutation = (1 2)(2 3 5)(3 4 5 6).
Solution: has order 4.
The permutation is dened as a product of cycles that are not disjoint. To nd the order of , we
need to represent as a
MATH 433100
June 18, 2010
Quiz 9
Problem 1. Express the permutation = (1 2 4)(2 3 4 5) as a product of transpositions and
nd the sign of .
Solution: = (1 2)(2 4)(2 3)(3 4)(4 5), sgn( ) = 1.
We have (1 2 4) = (1 2)(2 4) and (2 3 4 5) = (2 3)(3 4)(4 5), hen
MATH 433100
June 22, 2010
Quiz 10: Solutions
nk
, where n and k are
0n
integers. Under the operation of matrix multiplication, does this set form a semigroup? Does M
form a group? Is the operation commutative on M ? Explain.
Problem 1.
Let M be the set of
MATH 433100
June 23, 2010
Quiz 11: Solutions
n0
, where n and k are
kn
integers. Under the operations of matrix addition and multiplication, does this set form a ring? Does
M form a eld? Explain.
Problem 1.
Let M be the set of all 2 2 matrices of the form
MATH 433100
June 24, 2010
Quiz 12: Solutions
Problem 1. Consider the multiplicative group GL(2, Z2 ) of 2 2 matrices with entries from
the eld Z2 and nonzero determinant. How many elements are there in this group?
Solution: The group GL(2, Z2 ) consists o