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... classes. If a ∈ Z then let ā denote the set of integers congruent to a modulo m, that is ā = {n ∈ Z|n ≡ a (mod m)}. This is the same as saying ā is the set of integers of the form a + km (k ∈ Z). Definition 3.3. A set of the form ā is called a congruence class modulo m. Definition 3.4. The set o ...
... classes. If a ∈ Z then let ā denote the set of integers congruent to a modulo m, that is ā = {n ∈ Z|n ≡ a (mod m)}. This is the same as saying ā is the set of integers of the form a + km (k ∈ Z). Definition 3.3. A set of the form ā is called a congruence class modulo m. Definition 3.4. The set o ...
On the number of parts of integer partitions lying in given residue
... Several authors [6, 7, 8, 9], have also addressed the question how many parts with certain properties, e.g. congruence conditions, there are asymptotically in a given partition of a positive integer n. This also the object of the present paper. Before proceeding, let us introduce and fix some notati ...
... Several authors [6, 7, 8, 9], have also addressed the question how many parts with certain properties, e.g. congruence conditions, there are asymptotically in a given partition of a positive integer n. This also the object of the present paper. Before proceeding, let us introduce and fix some notati ...
Math 1530 Final Exam Spring 2013 Name:
... multiple if both a and b divide m, and if m0 is any other element divisible by both a and b then m divides m0 . If R is a PID, prove that a least common multiple always exists. Solution. There are (at least) three ways to prove this. First, translating the LCM property into the language of ideals, i ...
... multiple if both a and b divide m, and if m0 is any other element divisible by both a and b then m divides m0 . If R is a PID, prove that a least common multiple always exists. Solution. There are (at least) three ways to prove this. First, translating the LCM property into the language of ideals, i ...
Section 4.3 - math-clix
... the remainder. The quotient must have degree less than that of the dividend, P(x). The remainder must be either 0 or have degree less than that of the divisor. P(x) = d(x) • Q(x) + R(x) ...
... the remainder. The quotient must have degree less than that of the dividend, P(x). The remainder must be either 0 or have degree less than that of the divisor. P(x) = d(x) • Q(x) + R(x) ...
Review guide for Exam 2
... §1.5: In this section, we learned how to solve linear congruences, that is congruences of the form ax ≡ b (mod n) for fixed a, b, n, as well as systems of several linear congruences. For this section, your goal is to be able to efficiently solve congruences which I give to you explicitly, so make su ...
... §1.5: In this section, we learned how to solve linear congruences, that is congruences of the form ax ≡ b (mod n) for fixed a, b, n, as well as systems of several linear congruences. For this section, your goal is to be able to efficiently solve congruences which I give to you explicitly, so make su ...
Answers to some typical exercises
... remainder 0 when divided by 6, then we have done. If none of them have remainder 0, then there are at most 5 cases (pigeonhole) of the remainder. Thus, at least two of them must have the same remainder. The positive difference of these two is a subsequence whose sum is divisible by 6. ...
... remainder 0 when divided by 6, then we have done. If none of them have remainder 0, then there are at most 5 cases (pigeonhole) of the remainder. Thus, at least two of them must have the same remainder. The positive difference of these two is a subsequence whose sum is divisible by 6. ...
Lecture 11: the Euler φ-function In the light of the previous lecture
... Consider the number a b + ab . I claim that it is coprime to ab. To see what let p | ab and p | a′ b + ab′ . By Gauss’s lemma, since a and b are coprime we have that p | a or p | b. Suppose the former, without loss of generality. Then it is easy to see that this implies that p | a′ b. Now a′ and a a ...
... Consider the number a b + ab . I claim that it is coprime to ab. To see what let p | ab and p | a′ b + ab′ . By Gauss’s lemma, since a and b are coprime we have that p | a or p | b. Suppose the former, without loss of generality. Then it is easy to see that this implies that p | a′ b. Now a′ and a a ...
Threshold in N(n,p)
... interesting Suduko puzzle one has to erase enough numbers of the solution in order to keep the solution unique and possible. There is a threshold on the number of integers that can be erased and the puzzle is still solvable. ...
... interesting Suduko puzzle one has to erase enough numbers of the solution in order to keep the solution unique and possible. There is a threshold on the number of integers that can be erased and the puzzle is still solvable. ...
Remainder Theorem
... Factor Theorem If the remainder f(r) = R = 0, then (x-r) is a factor of f(x). The Factor Theorem is powerful because it can be used to find the roots of polynomial equations. Example 3: Is x 4 a factor of 3x 3 x 2 20 x 5 ? For this question we need to find out if dividing 3x 3 x 2 20 x ...
... Factor Theorem If the remainder f(r) = R = 0, then (x-r) is a factor of f(x). The Factor Theorem is powerful because it can be used to find the roots of polynomial equations. Example 3: Is x 4 a factor of 3x 3 x 2 20 x 5 ? For this question we need to find out if dividing 3x 3 x 2 20 x ...
Algebraic Fractions
... To multiply fractions, cancel down any factors, and then multiply the numerators and the denominators. To divide, change into a multiplication by taking the reciprocal of the divisor. We can only add/subtract fractions when they have the same denominator, so change every term so that it has the same ...
... To multiply fractions, cancel down any factors, and then multiply the numerators and the denominators. To divide, change into a multiplication by taking the reciprocal of the divisor. We can only add/subtract fractions when they have the same denominator, so change every term so that it has the same ...
Full text
... A Niven number is a positive integer that is divisible by its digital sum. That is, if n is an integer and s(n) Niven number if and only if sin) ...
... A Niven number is a positive integer that is divisible by its digital sum. That is, if n is an integer and s(n) Niven number if and only if sin) ...
MA109, Activity 31: Dividing Polynomials (Section 4.2, pp. 325
... Homework (Sec. 4.2): # 1,3,5,11,13,19,22,27,31,36,43,53 (pp. 331-332). Reading for next lecture: Sec. 4.3 (pp. 333-340). ...
... Homework (Sec. 4.2): # 1,3,5,11,13,19,22,27,31,36,43,53 (pp. 331-332). Reading for next lecture: Sec. 4.3 (pp. 333-340). ...
Mat2225, Number Theory, Homework 1
... 1. Use the Euclidean Algorithm inductively to find numbers a, b, c ∈ Z such that 30a + 105b + 42c = gcd(30, 105, 42) Also, use the method of prime-factorization to confirm your answer for the greatest common divisor gcd(30, 105, 42). Calculate lcm(30, 105, 42) as well. 2. Prove that for any a, b ∈ Z ...
... 1. Use the Euclidean Algorithm inductively to find numbers a, b, c ∈ Z such that 30a + 105b + 42c = gcd(30, 105, 42) Also, use the method of prime-factorization to confirm your answer for the greatest common divisor gcd(30, 105, 42). Calculate lcm(30, 105, 42) as well. 2. Prove that for any a, b ∈ Z ...
Assignment 4 Solutions - Math @ McMaster University
... x, x2 = 3, x3 = 3x, x4 = −1. This implies that x8 = 1. Since x6 = x2 · x4 = −3 = 2, we are done: The order of x is 8. We have (x + 2)2 = x2 + 4x + 4 = 4x + 2, (x + 2)3 = (4x + 2)(x + 2) = 4x2 + 4 = 1, which shows that x + 2 has order 3. b) We note that 3 + 2x = x · (x + 2). The order of 3 + 2x is th ...
... x, x2 = 3, x3 = 3x, x4 = −1. This implies that x8 = 1. Since x6 = x2 · x4 = −3 = 2, we are done: The order of x is 8. We have (x + 2)2 = x2 + 4x + 4 = 4x + 2, (x + 2)3 = (4x + 2)(x + 2) = 4x2 + 4 = 1, which shows that x + 2 has order 3. b) We note that 3 + 2x = x · (x + 2). The order of 3 + 2x is th ...
Chapter 2 Algebra Review 2.1 Arithmetic Operations
... Result 4 (The Binomial Theorem) If n is a positive integer, then (a + b)n = an + nan 1b + ...
... Result 4 (The Binomial Theorem) If n is a positive integer, then (a + b)n = an + nan 1b + ...
THE DIVISOR PROBLEM ON SQUARE
... . Under the Riemann hypothesis, by the elementary method, from (2) we can not change the error term of . But to our surprise, following the proof of Theorem we can reduce the error term to ...
... . Under the Riemann hypothesis, by the elementary method, from (2) we can not change the error term of . But to our surprise, following the proof of Theorem we can reduce the error term to ...
Definition of the Quadratic Formula
... While factoring may not always be successful, the Quadratic Formula can always find the solution. For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the squar ...
... While factoring may not always be successful, the Quadratic Formula can always find the solution. For the Quadratic Formula to work, you must have your equation arranged in the form "(quadratic) = 0". Also, the "2a" in the denominator of the Formula is underneath everything above, not just the squar ...
ADDING AND COUNTING Definition 0.1. A partition of a natural
... Bootstrapping the ideas created by Hardy and Ramanujan, Rademacher was able to write down an exact formula for p(n). Theorem 0.5 (Rademacher (1943)). For each n ∈ N we have ...
... Bootstrapping the ideas created by Hardy and Ramanujan, Rademacher was able to write down an exact formula for p(n). Theorem 0.5 (Rademacher (1943)). For each n ∈ N we have ...
Intermediate Algebra Section 5.3 – Dividing Polynomials
... To divide a polynomial by a polynomial other than a monomial, we use a method called long division. This method is similar to long division of real numbers. When using long division for polynomials, the polynomials must be written in descending order. Example: ...
... To divide a polynomial by a polynomial other than a monomial, we use a method called long division. This method is similar to long division of real numbers. When using long division for polynomials, the polynomials must be written in descending order. Example: ...
2. Are the following polynomials irreducible over Q? (a) 3 x + 18 x +
... irreducible in Z2 [x]. This implies that the polynomial is irreducible over Q. 3. Factor x5-10x4+24x3+9x2-33x-12 over Q. Solution: The possible rational roots of f(x) = x5-10x4+24x3+9x2-33x-12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12. We have f(1) = 21, so for any root we must have (r-1) | 21, so this elim ...
... irreducible in Z2 [x]. This implies that the polynomial is irreducible over Q. 3. Factor x5-10x4+24x3+9x2-33x-12 over Q. Solution: The possible rational roots of f(x) = x5-10x4+24x3+9x2-33x-12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12. We have f(1) = 21, so for any root we must have (r-1) | 21, so this elim ...
b - Stony Brook Mathematics
... has a solution. More over, any two solutions are congruent modulo mn. 2) Proof of Chinese remainder theorem Since (n,m)=1, there exist integers r and s such that rm+sn=1. Then rm ≡1 (modn) and sn ≡1 (modm). Let x=arm+bsn. Then a direct computation verifies that x ≡arm ≡ a (modn) and x ≡ bsn b ≡ (mod ...
... has a solution. More over, any two solutions are congruent modulo mn. 2) Proof of Chinese remainder theorem Since (n,m)=1, there exist integers r and s such that rm+sn=1. Then rm ≡1 (modn) and sn ≡1 (modm). Let x=arm+bsn. Then a direct computation verifies that x ≡arm ≡ a (modn) and x ≡ bsn b ≡ (mod ...
Full text
... (4) in November 1947, but at that time did not see its application to the Fibonacci and similar sequences and therefore withheld its publication. P. S. Bruckman independently, and recently, noticed the more general formula (14). Alternate methods of proof appear in [ 7 ] . ...
... (4) in November 1947, but at that time did not see its application to the Fibonacci and similar sequences and therefore withheld its publication. P. S. Bruckman independently, and recently, noticed the more general formula (14). Alternate methods of proof appear in [ 7 ] . ...
Chinese remainder theorem
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The Chinese remainder theorem is a result about congruences in number theory and its generalizations in abstract algebra. It was first published in the 3rd to 5th centuries by the Chinese mathematician Sun Tzu.In its basic form, the Chinese remainder theorem will determine a number n that, when divided by some given divisors, leaves given remainders. For example, what is the lowest number n that when divided by 3 leaves a remainder of 2, when divided by 5 leaves a remainder of 3, and when divided by 7 leaves a remainder of 2?