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
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 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 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 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 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 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 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 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