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 = [1]37 . Solution: Since gcd(5, 37) = 1, there is
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 the sum of the digits of a AND 2 divides the units digit
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 units in Z17 , is a eld of 16 elements.
However, this
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 element
a F11 , i.e. the generator polynomial is of the
The Great Gatsby
By F. Scott Fitzgerald
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Then wear the gold hat, if that will move
her; If you can bounce high, bounce for
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 remainder of 115, which is 5 23.
So, m = 5, 23, 115 would
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 ) = 3 (Check!), and umber of possible words = 81. Since
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 ker(G), i.e. f(g.g ) =
f(g 1 )= e.
But f(g.g ) = f(g).f
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 =
(a(m) )k a 1 a(mod m) by Euler.
9C.62[graded out of 5
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 algorithm, for instance).
3. Computing powers of 2 mod 17 w
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 using.
SHOW ALL OF YOUR WORK AND JUSTIFY YOUR ANSWERS.
1
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 numbers. (What are the roots?)
(2) Problem 18 on page 301.
So
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 is 3; and when counted by sevens the remainder is 2.
How
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 own inverse. Hence this is a non-cyclic subgroup of U16
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 characteristic 2, then we know 1 + 1 = 2 = 0. So by the distr
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 on page 130.
Solution: We know that 1 + 0 = 1 b (1 + 0)
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 n = pq, where p is the smallest prime dividing n. Assu
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 = 2.
Inductive Step: Suppose that this statement holds fo
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 as you show clearly what you are using. You must EXPLAIN
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 are
using. SHOW ALL OF YOUR WORK AND JUSTIFY YOUR ANSWER
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).
So the general solution is x = 1 + 23k , where k is
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 as you show clearly what you are using. You must EXPLAIN
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 to the following minimal polynomial p(x) = x4 10x + 1.
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_1, 3, 7, 9. Clearly 1 = cfw_1, 3 = cfw_3, 32 , 33 , 34