MATH10040 Chapter 3: Congruences and the Chinese Remainder
... Since 20 = 4 · 5 and (4, 5) = 1, by Corollary 3.3, it is enough to show that 871003 ≡ 3 (mod 4) and 871003 ≡ 3 (mod 5). Now 87 ≡ −1 (mod 4) =⇒ 871003 ≡ (−1)1003 ≡ −1 ≡ 3 (mod 4). 87 ≡ 2 (mod 5) =⇒ 874 ≡ 24 ≡ 1 (mod 5). Thus 871000+3 ≡ 21000 · 23 ≡ 23 ≡ 3 (mod 5). Example 3.5. Find the last two digit ...
... Since 20 = 4 · 5 and (4, 5) = 1, by Corollary 3.3, it is enough to show that 871003 ≡ 3 (mod 4) and 871003 ≡ 3 (mod 5). Now 87 ≡ −1 (mod 4) =⇒ 871003 ≡ (−1)1003 ≡ −1 ≡ 3 (mod 4). 87 ≡ 2 (mod 5) =⇒ 874 ≡ 24 ≡ 1 (mod 5). Thus 871000+3 ≡ 21000 · 23 ≡ 23 ≡ 3 (mod 5). Example 3.5. Find the last two digit ...
MATH10040 Chapter 3: Congruences and the Chinese Remainder
... Remark 1.5. Observe that if m ≥ 1 and a ∈ Z, then the set of numbers which are congruent to a modulo m are the numbers in the list . . . , a − 2m, a − m, a, a + m, a + 2m, a + 3m, . . . This is a (doubly) infinite sequence of numbers. The difference between successive terms is always m. (Thus this i ...
... Remark 1.5. Observe that if m ≥ 1 and a ∈ Z, then the set of numbers which are congruent to a modulo m are the numbers in the list . . . , a − 2m, a − m, a, a + m, a + 2m, a + 3m, . . . This is a (doubly) infinite sequence of numbers. The difference between successive terms is always m. (Thus this i ...
Notes for Number theory (Fall semester)
... The older term for number theory is arithmetic. By the early twentieth century, it had been superseded by ”number theory”. The word ”arithmetic” (from the Greek, arithmos which means ”number”) is used by the general public to mean ”elementary calculations”; it has also acquired other meanings in mat ...
... The older term for number theory is arithmetic. By the early twentieth century, it had been superseded by ”number theory”. The word ”arithmetic” (from the Greek, arithmos which means ”number”) is used by the general public to mean ”elementary calculations”; it has also acquired other meanings in mat ...
Lectures on Number Theory
... of ideals is an ideal.) By definition, an integer x belongs to this intersection if and only if a|x and b|x, i.e. if and only if x is a common multiple of a and b. Thus, the ideal aZ ∩ bZ coincides with the set of all common multiples of the numbers a and b. This observation leads us to the followin ...
... of ideals is an ideal.) By definition, an integer x belongs to this intersection if and only if a|x and b|x, i.e. if and only if x is a common multiple of a and b. Thus, the ideal aZ ∩ bZ coincides with the set of all common multiples of the numbers a and b. This observation leads us to the followin ...
Midterm #3: practice
... (b) We need to compute 2340 (mod 341). We proceed using binary exponentiation as in the previous part. The values we get modulo 341 are: 22 = 4, 24 = 16, 28 = 256, 216 = 64, 232 = 4, so that, again, values repeat. In the end, we nd that 2340 1 (mod 341). This means that 341 is a pseudoprime to th ...
... (b) We need to compute 2340 (mod 341). We proceed using binary exponentiation as in the previous part. The values we get modulo 341 are: 22 = 4, 24 = 16, 28 = 256, 216 = 64, 232 = 4, so that, again, values repeat. In the end, we nd that 2340 1 (mod 341). This means that 341 is a pseudoprime to th ...
nicely typed notes
... Examples. Note that if m | (a − b), then m | (b − a), since b − a = −(a − b). • 8 and 5 have the same remainder when divided by 3 since 8−5 = 3 and 3 | 3, so 3 | (8 − 5). • 13 and 25 have the same remainder when divided by 6 since 13 − 25 = −12 and 6 | −12, so 6 | (13 − 25). • 27 and −25 have the s ...
... Examples. Note that if m | (a − b), then m | (b − a), since b − a = −(a − b). • 8 and 5 have the same remainder when divided by 3 since 8−5 = 3 and 3 | 3, so 3 | (8 − 5). • 13 and 25 have the same remainder when divided by 6 since 13 − 25 = −12 and 6 | −12, so 6 | (13 − 25). • 27 and −25 have the s ...
Lecture 5 The Euclidean Algorithm
... To calculate Q(x) and R(x) it suffices to find R(x) since we can divide A(x)- R(x) by B(x) to get R(x) The uniqueness of the remainder says if in any way you arrange to write A(x) = B(x)K(x) + P(x) where P(x) is zero or of smaller degree than B(x) then it must be that P(x) is the R(x) you would get ...
... To calculate Q(x) and R(x) it suffices to find R(x) since we can divide A(x)- R(x) by B(x) to get R(x) The uniqueness of the remainder says if in any way you arrange to write A(x) = B(x)K(x) + P(x) where P(x) is zero or of smaller degree than B(x) then it must be that P(x) is the R(x) you would get ...
on pairwise hyperconnected spaces
... Let (X ) be pairwise hyperconnected, , be hyperconnected and be irresolvable. Let A, B 2 SO( ) SO( )] n fg. Case 1. Let A, B 2 SO( ) n fg. Then from Theorem 2.1, A \ B 2 SO( ) n fg. Case 2. Let A, B 2 SO( ) n fg. Then from Corollary 2.1, A \ B 2 SO( ) n fg. Case 3. Le ...
... Let (X ) be pairwise hyperconnected, , be hyperconnected and be irresolvable. Let A, B 2 SO( ) SO( )] n fg. Case 1. Let A, B 2 SO( ) n fg. Then from Theorem 2.1, A \ B 2 SO( ) n fg. Case 2. Let A, B 2 SO( ) n fg. Then from Corollary 2.1, A \ B 2 SO( ) n fg. Case 3. Le ...
15(1)
... Mathematics Department, University of Santa Clara, Santa Clara, California 9 5 0 5 3 . All checks ($15.00 per year) should be made out to the Fibonacci Association or The Fibonacci Quarterly. Two copies of manuscripts intended for publication in the Quarterly should be sent to Verner E. Hoggatt, Jr. ...
... Mathematics Department, University of Santa Clara, Santa Clara, California 9 5 0 5 3 . All checks ($15.00 per year) should be made out to the Fibonacci Association or The Fibonacci Quarterly. Two copies of manuscripts intended for publication in the Quarterly should be sent to Verner E. Hoggatt, Jr. ...
Solutions
... m ∈ Z. We get a positive solution smaller than 200 only for m = 0, 1, so 88 and 193 are the only solutions to our problem. Problem 3. a) Define (a, b). Using Euclid’s algorithm compute (889, 168) and find x, y ∈ Z such that (889, 168) = x · 889 + y · 168 (check your answer!). b) Let a be an integer. ...
... m ∈ Z. We get a positive solution smaller than 200 only for m = 0, 1, so 88 and 193 are the only solutions to our problem. Problem 3. a) Define (a, b). Using Euclid’s algorithm compute (889, 168) and find x, y ∈ Z such that (889, 168) = x · 889 + y · 168 (check your answer!). b) Let a be an integer. ...
109_lecture4_fall05
... To calculate Q(x) and R(x) it suffices to find R(x) since we can divide A(x)- R(x) by B(x) to get R(x) The uniqueness of the remainder says if in any way you arrange to write A(x) = B(x)K(x) + P(x) where P(x) is zero or of smaller degree than B(x) then it must be that P(x) is the R(x) you would get ...
... To calculate Q(x) and R(x) it suffices to find R(x) since we can divide A(x)- R(x) by B(x) to get R(x) The uniqueness of the remainder says if in any way you arrange to write A(x) = B(x)K(x) + P(x) where P(x) is zero or of smaller degree than B(x) then it must be that P(x) is the R(x) you would get ...
Number Theory Notes
... Example. Find 5−1 modulo 9. At the moment, the only way we have of working this is out is trial and error. Thankfully, we don’t have to look far: 5 ∗ 2 = 10 ≡ 1 (mod 9). Thus, 5−1 ≡ 2 (mod 9). Note that it is incorrect to say that 5−1 = 2 modulo 9, as it is not the only solution to the problem. By ...
... Example. Find 5−1 modulo 9. At the moment, the only way we have of working this is out is trial and error. Thankfully, we don’t have to look far: 5 ∗ 2 = 10 ≡ 1 (mod 9). Thus, 5−1 ≡ 2 (mod 9). Note that it is incorrect to say that 5−1 = 2 modulo 9, as it is not the only solution to the problem. By ...
Modular Arithmetic
... respectively. On the first step, we find the inverses of each modulo with respect to each later modulo in the list. This can be done by the extended Euclidean algorithm (see the proof of the theorem for solving linear congruence.) ...
... respectively. On the first step, we find the inverses of each modulo with respect to each later modulo in the list. This can be done by the extended Euclidean algorithm (see the proof of the theorem for solving linear congruence.) ...
SOLUTIONS OF SOME CLASSES OF CONGRUENCES Eugen
... Proof. The proof of Proposition 2 is given by the following chain of formulas: xkl ≡ al (mod p), xkl+2 ≡ x2 al (mod p), x2 al ≡ 1 (mod p), x2 ≡ a−l (mod p). Actually, xkl+2 ≡ 1 (mod p) by Fermat theorem. If (5) has a solution, then it has two solutions. It is necessary to check whether they are the ...
... Proof. The proof of Proposition 2 is given by the following chain of formulas: xkl ≡ al (mod p), xkl+2 ≡ x2 al (mod p), x2 al ≡ 1 (mod p), x2 ≡ a−l (mod p). Actually, xkl+2 ≡ 1 (mod p) by Fermat theorem. If (5) has a solution, then it has two solutions. It is necessary to check whether they are the ...
*7. Polynomials
... look at factorization, or dividing, it is best to restrict to sets F in which we can find inverses. (We can talk of dividing in Q, R or C but not in Zp . Instead we have to remember that to divide by something is to multiply by it’s inverse.) (Aside: In Z6 does [3]6 have an inverse? i.e. does there ...
... look at factorization, or dividing, it is best to restrict to sets F in which we can find inverses. (We can talk of dividing in Q, R or C but not in Zp . Instead we have to remember that to divide by something is to multiply by it’s inverse.) (Aside: In Z6 does [3]6 have an inverse? i.e. does there ...
Polynomial Resultants - University of Puget Sound
... Proof. To see this, suppose γ is a root shared by both polynomials, then one term of the product is (γ − γ) = 0, hence the whole product vanishes. Conversely, suppose Res(f, g, x) = 0, then, since F is a field and therefore has no zero divisors, at least one term of the product zero. Suppose that te ...
... Proof. To see this, suppose γ is a root shared by both polynomials, then one term of the product is (γ − γ) = 0, hence the whole product vanishes. Conversely, suppose Res(f, g, x) = 0, then, since F is a field and therefore has no zero divisors, at least one term of the product zero. Suppose that te ...
Undergrad covering talk - Dartmouth Math Home
... B0 so that for B ≥ B0 there is no covering with distinct moduli from [B, KB]. Conjecture (Erdős, Graham). For each K > 1, there are dK > 0, B0 such that for B ≥ B0 and for any congruences with distinct moduli from [B, KB], at least density dK of Z remains uncovered. ...
... B0 so that for B ≥ B0 there is no covering with distinct moduli from [B, KB]. Conjecture (Erdős, Graham). For each K > 1, there are dK > 0, B0 such that for B ≥ B0 and for any congruences with distinct moduli from [B, KB], at least density dK of Z remains uncovered. ...
Sample pages 2 PDF
... The next ideas that are necessary are the concepts of greatest common divisor and least common multiple. Definition 2.2.2 Given nonzero integers a, b their greatest common divisor or GCD d > 0 is a positive integer which is a common divisor, that is, d|a and d|b, and if d1 is any other common diviso ...
... The next ideas that are necessary are the concepts of greatest common divisor and least common multiple. Definition 2.2.2 Given nonzero integers a, b their greatest common divisor or GCD d > 0 is a positive integer which is a common divisor, that is, d|a and d|b, and if d1 is any other common diviso ...
UNIT_11
... units. Also, they are familiar with the concept of numbers. They also know factorization of numbers into prime factors in arithmetic. However, a brief recapitulation by the teacher can help them to revise their basic concepts before you start actual teaching of this unit. The concepts to be learnt i ...
... units. Also, they are familiar with the concept of numbers. They also know factorization of numbers into prime factors in arithmetic. However, a brief recapitulation by the teacher can help them to revise their basic concepts before you start actual teaching of this unit. The concepts to be learnt i ...
Decimal expansions of fractions
... is an exact correspondence between the numbers modulo n and pairs modulo p and q. I’ll give two proofs of this, one short, the other longer but constructive. The first proof shows directly that there are no repeats of pairs. Suppose m 1 and m2 correspond to the same pair a, b). Then m1 = a modulo ; ...
... is an exact correspondence between the numbers modulo n and pairs modulo p and q. I’ll give two proofs of this, one short, the other longer but constructive. The first proof shows directly that there are no repeats of pairs. Suppose m 1 and m2 correspond to the same pair a, b). Then m1 = a modulo ; ...
Better polynomials for GNFS - Mathematical Sciences Institute, ANU
... but since they are not published elsewhere, we mention them for completeness. We want to produce a polynomial with small L2 -norm by translation and rotation. We focus on degree 6 here for sake of clarity, it is straightforward to generalize to other degrees. ...
... but since they are not published elsewhere, we mention them for completeness. We want to produce a polynomial with small L2 -norm by translation and rotation. We focus on degree 6 here for sake of clarity, it is straightforward to generalize to other degrees. ...
Better Polynomials for GNFS
... but since they are not published elsewhere, we mention them for completeness. We want to produce a polynomial with small L2 -norm by translation and rotation. We focus on degree 6 here for sake of clarity, it is straightforward to generalize to other degrees. In the raw polynomial, c0 , c1 , c2 , c3 ...
... but since they are not published elsewhere, we mention them for completeness. We want to produce a polynomial with small L2 -norm by translation and rotation. We focus on degree 6 here for sake of clarity, it is straightforward to generalize to other degrees. In the raw polynomial, c0 , c1 , c2 , c3 ...
THE CHINESE REMAINDER THEOREM INTRODUCED IN A
... have collected a pile of coconuts which they plan to divide equally among themselves the next morning. Not trusting the others, one pirate wakes up during the night and divides the coconuts into ve equal parts with one left over, which he gives to the monkey. The pirate then hides his portion of th ...
... have collected a pile of coconuts which they plan to divide equally among themselves the next morning. Not trusting the others, one pirate wakes up during the night and divides the coconuts into ve equal parts with one left over, which he gives to the monkey. The pirate then hides his portion of th ...
Full text
... ally, we define a complete Fibonacci system modulo m to be a maximal set of pairwise inequivalent Fibonacci cycles modulo m. Note that the total number of terms appearing in such a system is m 2 . The idea behind this definition is simple; it is a compact way of r e p r e senting all possible Fibona ...
... ally, we define a complete Fibonacci system modulo m to be a maximal set of pairwise inequivalent Fibonacci cycles modulo m. Note that the total number of terms appearing in such a system is m 2 . The idea behind this definition is simple; it is a compact way of r e p r e senting all possible Fibona ...
Chinese remainder theorem
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?