Mathematics 336
5 problems, 20 points each
Prelim 1 Solutions
February 22, 2008
1. Solve, if possible, each of the following. If not possible, state why. (a) Find an integer b so that [5]37 [b]37
Homework 2: Math 3360 Applicable Algebra
Chapter 5: Congruence
24. Find a divisibility test for 6.
Solution: Combine the divisibility tests for 2 and 3, i.e. a number a is divisible by 6 i 3
divides t
MATH 3360
Sketch of solutions for Assignment 10
17C.25
+ x2 )
1
(3x
2
19A.10
There are many possible solutions, for instance 2, 2 = 59, etc.
Additional Problem 1
Some students wrote U17 , the group of
MATH 3360
Sketch of solutions for Assignment 12
Ch.3 4.1
Work on the ring F11 [x]/(x11 1), and for the RS code to be 3-error correcting,
you need a chain of 3 2 = 6 consecutive powers of a primitive e
The Great Gatsby
By F. Scott Fitzgerald
Download free eBooks of classic literature, books and
novels at Planet eBook. Subscribe to our free eBooks
blog and email newsletter.
Then wear the gold hat, if
MATH 3360
Sketch of solutions for Assignment 9
14A.5
(i) Using root theorem the answer is no in Q[x]. The answer is no in Z[x]
since it was no in Q[x].
(ii) Dividing x4 + x3 + x + 4 by x 3 gives a rem
MATH 3360
Solutions for Selected Porblems in Assignment 8
CS 2.12
One can use the sphere-packing bound equation to see if the code is perfect.
However, a more intuitive way is this:
|C | = 9, d(C ) =
Selected Solutions for HW 6
32. Let f: G G be a group homomorphism. Then ker(f) is a subgroup of G, and f is
1-1 iff ker(f) = cfw_e.
Proof:
Let g,g be in ker(G). We need to show g.g , g 1 are both in
MATH 3360
Sketch of solutions for Assignment 5
9C.49
Since (18) = 6, we have 56 = 5 55 1(mod 18). Answer: 55 .
9C.50[graded out of 2 points]
f 1(mod (m) f = k(m) + 1 for some k Z. Thus, af = ak(m)+1 =
Brief solutions to Prelim 2
1. A binary code of dimension k has 2k codewords. So the answer is 2.
2. The multiplicative inverse of x3 + x + 1 mod x4 + x3 + 1 in F2 [x] is x3 + x (by
the Euclidean algo
Mathematics 3360
Prelim 2
April 15, 2010
No books, notes, or electronic devices may be used. You may use anything that has
been given in class or in the book as long as you show clearly what you are u
MATH 3360 Applicable Algebra. HW 9 Solutions
(1) Problem 11 on page 298.
Solution: One example can be R = Z/mZ, f ( x) = x2 . For a given n, we let m be a product of
squares of n distinct prime number
Homework 8: Math 3360 Applicable Algebra
Chapter 12: Chinese Remainder Theorem
19. There is an unknown number of objects. When counted by threes the remainder is 2;
when counted by ves the remainder i
MATH 3360 Applicable Algebra. HW 7 Solutions
(1) Problem 13 on page 230. (graded)
Solution: Consider H = cfw_1, 7, 9, 15. Note that this set is closed under the operation and that each
element is its
Homework 6: Math 3360 Applicable Algebra
Chapter 7: Rings and Fields
44. Let F be a eld of characteristic 2. Show that for any a, b F
(i) a = a
(ii) (a + b)2 = a2 + b2
Solution: (i) Since F has charac
MATH 3360 Applicable Algebra. HW 5 Solutions
(1) Problem 2 on page 130.
Solution: By denition (a) is the inverse of (a). Since a + (a) = 0 and inverses are unique,
we must have (a) = a.
(2) Problem 7
MATH 3360 Applicable Algebra. HW 3 Solutions.
(1) Problem 5 on page 56. Show that if n is not a prime or a square of a prime,
then n has a prime factor smaller than the square root of n.
Solution: Let
Homework 1: Math 3360 Applicable Algebra
p.13
3. Claim: Show that 1 + 2 + 22 + . + 2n1 = 2n1 for every n > 1.
Pf: Base case: Let n = 2. Then we have 1 + 2 = 3 = 22 1, so the statement is true for n =
Mathematics 336
4 problems, 25 points each
Prelim 2
April 4, 2008
No books, notes or electronic devices may be used. Your proofs may use anything that has been given in class or in the book, as long a
Mathematics 3360
Prelim 1
February 23, 2010
No books, notes, or electronic devices may be used. You may use anything that
has been given in class or in the book, as long as you show clearly what you a
Brief solutions to Prelim 1
1. Yes, because (11, 41) = 1. Using the Euclidean algorithm, we nd that the
inverse is [15].
2. Using the cancellation laws, we have
9x 9
(mod 69) 3x 3
(mod 23) x 1
(mod 23
Mathematics 336
4 problems, 25 points each
Prelim 2
April 4, 2008
No books, notes or electronic devices may be used. Your proofs may use anything that has been given in class or in the book, as long a
MATH 3360 Applicable Algebra. HW 14 Solutions
Problem 24.4
Let x = 2 + 3, then x2 = 5 + 2 6. Subtracting 5 from both sides
and squaring, gives x4 10x + 25 = (x2 5)2 = 4 6 = 24. So 2 + 3
is a solution
Homework 13: Math 3360 Applicable Algebra
Chapter 19: Cyclic Groups and Cryptography
15. Show that the group U10 of units of Z/10Z is cyclic. Find a generator of U1 0.
Solution: Here we have U10 = cfw