Solution
... No matter what, we have shown that 2 raised to the tenth power is congruent to 1 modulo 11. If we divide the exponent 11, 213 by 10 we obtain 11, 213 = 10 · 1, 121 + 3. Thus 211,213 = 210·1,121+3 = (210 )1,121 · 23 ≡ (1)1,121 · 23 ≡ 1 · 8 ≡ 8 (mod 11). Thus, we have shown that 211,213 ≡ 8 (mod 11), ...
... No matter what, we have shown that 2 raised to the tenth power is congruent to 1 modulo 11. If we divide the exponent 11, 213 by 10 we obtain 11, 213 = 10 · 1, 121 + 3. Thus 211,213 = 210·1,121+3 = (210 )1,121 · 23 ≡ (1)1,121 · 23 ≡ 1 · 8 ≡ 8 (mod 11). Thus, we have shown that 211,213 ≡ 8 (mod 11), ...
How to solve f (x ) = g (y )...
... (Observed by Goormaghtigh nearly a century ago). However, it is still unknown whether there are only finitely many solutions in x, y , m, n. In fact, no other solutions are known. For any fixed bases x, y , it was proved only as recently as in 2002 that the number of solutions for m, n is at the mos ...
... (Observed by Goormaghtigh nearly a century ago). However, it is still unknown whether there are only finitely many solutions in x, y , m, n. In fact, no other solutions are known. For any fixed bases x, y , it was proved only as recently as in 2002 that the number of solutions for m, n is at the mos ...
Let m be a positive integer. Show that a mod m = b mod m if a ≡ b
... Since 3, 4, and 5 are pairwise relatively prime, we can use the Chinese Remainder Theorem. The answer will be unique module 3·4·5 = 60. Using the notation in the text, we have a1 = 2, m1 = 3, a2 = 1, m2 = 4, a3 = 3, m3 = 5 m = 60, M1 = 60/3 = 20, M2 = 60/4 = 15, M3 = 60/5 = 12. Then we need to find ...
... Since 3, 4, and 5 are pairwise relatively prime, we can use the Chinese Remainder Theorem. The answer will be unique module 3·4·5 = 60. Using the notation in the text, we have a1 = 2, m1 = 3, a2 = 1, m2 = 4, a3 = 3, m3 = 5 m = 60, M1 = 60/3 = 20, M2 = 60/4 = 15, M3 = 60/5 = 12. Then we need to find ...
Week 10
... Since we have checked all possible numbers of consecutive integers below five and none of them worked, the minimum number of consecutive integers required to produce a sum of 1000 is five. ...
... Since we have checked all possible numbers of consecutive integers below five and none of them worked, the minimum number of consecutive integers required to produce a sum of 1000 is five. ...
Section 4.3 - The Chinese Remainder Theorem
... Section 4.3 - The Chinese Remainder Theorem Exercise 4abc: Find all of the solutions to each system of linear congruences (a) x ≡ 4 (mod 11) x ≡ 3 (mod 17) ...
... Section 4.3 - The Chinese Remainder Theorem Exercise 4abc: Find all of the solutions to each system of linear congruences (a) x ≡ 4 (mod 11) x ≡ 3 (mod 17) ...
Math 301, Linear Congruences Linear
... We would like to multiply both sides of the congruence by an integer s that will “cancel” the 3. (So, in a certain sense, we are looking for a reciprocal for 3.) This means we want 3s ⌘ 1 mod 7. We list the multiples of 3 until we find one that is one more than a multiple of 7: 3, 6, 9, 12, 15. Sinc ...
... We would like to multiply both sides of the congruence by an integer s that will “cancel” the 3. (So, in a certain sense, we are looking for a reciprocal for 3.) This means we want 3s ⌘ 1 mod 7. We list the multiples of 3 until we find one that is one more than a multiple of 7: 3, 6, 9, 12, 15. Sinc ...
Fermat`s little theorem, Chinese Remainder Theorem
... If a and b are relatively prime if and only if there are integers m and n so that am + bn = 1. Working modulo b, this says Lemma 2. a and b are relatively prime if and only if there is an integer m such that am ≡ 1 ...
... If a and b are relatively prime if and only if there are integers m and n so that am + bn = 1. Working modulo b, this says Lemma 2. a and b are relatively prime if and only if there is an integer m such that am ≡ 1 ...
Lehmer`s problem for polynomials with odd coefficients
... with c2 = (log 5)/4 and cm = log( m2 + 1/2) for m > 2. We provide in Theorem 2.4 a characterization of polynomials f ∈ Z[x] for which there exists a polynomial F ∈ Dp with f | F and M(f ) = M(F ), where p is a prime number. The proof in fact specifies an explicit construction for such a polynomial F ...
... with c2 = (log 5)/4 and cm = log( m2 + 1/2) for m > 2. We provide in Theorem 2.4 a characterization of polynomials f ∈ Z[x] for which there exists a polynomial F ∈ Dp with f | F and M(f ) = M(F ), where p is a prime number. The proof in fact specifies an explicit construction for such a polynomial F ...
Week 7
... residues. There are (p−1)/2 even numbers between 1 and p−1 and there are (p−1)/2 quadratic residues. Thus the odd powers of g are the quadratic nonresidues. This shows that the quadratic residues are exactly the even powers of a primitive root g and the quadratic nonresidues are the odd powers of g. ...
... residues. There are (p−1)/2 even numbers between 1 and p−1 and there are (p−1)/2 quadratic residues. Thus the odd powers of g are the quadratic nonresidues. This shows that the quadratic residues are exactly the even powers of a primitive root g and the quadratic nonresidues are the odd powers of g. ...
Lecture 1-3: Abstract algebra and Number theory
... For g(x), h(x), g1 (x), h1 (x), s(x) ∈ F [x] we have (i) g(x) ≡ h(x) (mod f (x)) if and only if g(x) and h(x) leaves the same remainder when divided by f (x). (ii) g(x) ≡ g(x) (mod f (x)). (iii) If g(x) ≡ h(x) (mod f (x)) then h(x) ≡ g(x) (mod f (x)). iv) If g(x) ≡ h(x) (mod f (x)) and h(x) ≡ s(x) ( ...
... For g(x), h(x), g1 (x), h1 (x), s(x) ∈ F [x] we have (i) g(x) ≡ h(x) (mod f (x)) if and only if g(x) and h(x) leaves the same remainder when divided by f (x). (ii) g(x) ≡ g(x) (mod f (x)). (iii) If g(x) ≡ h(x) (mod f (x)) then h(x) ≡ g(x) (mod f (x)). iv) If g(x) ≡ h(x) (mod f (x)) and h(x) ≡ s(x) ( ...
ON SIMILARITIES BETWEEN EXPONENTIAL POLYNOMIALS AND
... function exp(− # ) on the interval (−∞, ∞) [25]. Now a collection of the basic properties of polynomials (# () will be listed. First one will be the following differential-difference equation (see (17)): ...
... function exp(− # ) on the interval (−∞, ∞) [25]. Now a collection of the basic properties of polynomials (# () will be listed. First one will be the following differential-difference equation (see (17)): ...
ELEMENTS OF NUMBER THEORY - Department of Mathematical
... 8. Prove that a 6–digit number of the form abcabc, a, b, c are the digits, is always divisible by 7,11, and 13. 9. There were seven sheets of paper. Some of them were cut into seven pieces. Some of the obtained pieces were cut again into seven pieces, and so on. At the end 1961 pieces altogether wer ...
... 8. Prove that a 6–digit number of the form abcabc, a, b, c are the digits, is always divisible by 7,11, and 13. 9. There were seven sheets of paper. Some of them were cut into seven pieces. Some of the obtained pieces were cut again into seven pieces, and so on. At the end 1961 pieces altogether wer ...
Enumeration in Algebra and Geometry
... The first topic is related to the classical question: “On how many pieces a certain collection of hyperplanes subdivides a linear space?” It is usually not hard to answer this question for a generic collection of hyperplanes. But for some special hyperplane arrangements the answer can be much more i ...
... The first topic is related to the classical question: “On how many pieces a certain collection of hyperplanes subdivides a linear space?” It is usually not hard to answer this question for a generic collection of hyperplanes. But for some special hyperplane arrangements the answer can be much more i ...
Divisibility
... Divisibility by 11 ( Study this method last – since it is not an obvious approach ) 1) Add every other digit beginning with the one’s digit ( one’s, hundred’s, ten-thousands,... ) 2) Add the remaining digits ( ten’s thousand’s, ... ) 3) If the first sum is less than the second one, add 11 to the fir ...
... Divisibility by 11 ( Study this method last – since it is not an obvious approach ) 1) Add every other digit beginning with the one’s digit ( one’s, hundred’s, ten-thousands,... ) 2) Add the remaining digits ( ten’s thousand’s, ... ) 3) If the first sum is less than the second one, add 11 to the fir ...
Grade 7/8 Math Circles Modular Arithmetic The Modulus Operator
... them to find remainders of large sums, differences, and multiplications. This is the study of modular arithmetic. Modular Addition Let’s say you wanted to find (130 + 41) mod 13. You could do this in two ways: • Add 130 to 41, then take its remainder modulo 13. ...
... them to find remainders of large sums, differences, and multiplications. This is the study of modular arithmetic. Modular Addition Let’s say you wanted to find (130 + 41) mod 13. You could do this in two ways: • Add 130 to 41, then take its remainder modulo 13. ...
Study Guide and Review
... Because r is an integer, 6r + 15 is an integer, and 3 k +1 (6r + 15) is divisible by 3. Therefore, 6 – 9 is divisible by 3. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true for n = 2, n = 3, and so on. ...
... Because r is an integer, 6r + 15 is an integer, and 3 k +1 (6r + 15) is divisible by 3. Therefore, 6 – 9 is divisible by 3. Because Pn is true for n = 1 and Pk implies Pk + 1, Pn is true for n = 2, n = 3, and so on. ...
Catalan Numbers, Their Generalization, and Their Uses
... see the historical note at the end of this ar~cle), which are closely related, both algebraically and conceptually, to the binomial coefficients. These Catalan numbers have many combinatorial interpretations, of which we will emphasize three, in addition to their interpretation in terms of 2-good pa ...
... see the historical note at the end of this ar~cle), which are closely related, both algebraically and conceptually, to the binomial coefficients. These Catalan numbers have many combinatorial interpretations, of which we will emphasize three, in addition to their interpretation in terms of 2-good pa ...
congruences modulo powers of 2 for the signature of complete
... Arf q' = Arf q + e(d)K F since e(K) is even and the Jacobi symbol (2 | d) = (-l)" W) . The proof in [10] is a simple counting argument. To deduce formula (2) note first that, since d is odd, Xx(d) is characteristic for X2(d) and Xt(d) • Xl(d) = d so (2) follows from Rokhlin's formula (16) if we show ...
... Arf q' = Arf q + e(d)K F since e(K) is even and the Jacobi symbol (2 | d) = (-l)" W) . The proof in [10] is a simple counting argument. To deduce formula (2) note first that, since d is odd, Xx(d) is characteristic for X2(d) and Xt(d) • Xl(d) = d so (2) follows from Rokhlin's formula (16) if we show ...
the arithmetical theory of linear recurring series
... repeats itself periodically, and that any one of its periods is a multiple of a certain least period which is called the characteristic number of (u) (or (a)) modulo m%. The number of non-repeating terms in (a) is called the numeric of (u) modulo m; if it is zero, (u) is said to be purely periodic^ ...
... repeats itself periodically, and that any one of its periods is a multiple of a certain least period which is called the characteristic number of (u) (or (a)) modulo m%. The number of non-repeating terms in (a) is called the numeric of (u) modulo m; if it is zero, (u) is said to be purely periodic^ ...
40(1)
... of individual articles (in whole or in part) provided complete reference is made to the source. Annual domestic Fibonacci Association membership dues, which include a subscription to THE FIBONACCI QUARTERLY, are $40 for Regular Membership, $50 for Library, $50 for Sustaining Membership, and $80 for ...
... of individual articles (in whole or in part) provided complete reference is made to the source. Annual domestic Fibonacci Association membership dues, which include a subscription to THE FIBONACCI QUARTERLY, are $40 for Regular Membership, $50 for Library, $50 for Sustaining Membership, and $80 for ...
M3P14 LECTURE NOTES 2: CONGRUENCES AND MODULAR
... Definition 1.1. Let n be a nonzero integer (usually taken to be positive) and let a and b be integers. We say a is congruent to b modulo n (written a ≡ b (mod n) ) if n | (a − b). For n fixed, it is easy to verify that congruence mod n is an equivalence relation, and therefore partitions Z into equi ...
... Definition 1.1. Let n be a nonzero integer (usually taken to be positive) and let a and b be integers. We say a is congruent to b modulo n (written a ≡ b (mod n) ) if n | (a − b). For n fixed, it is easy to verify that congruence mod n is an equivalence relation, and therefore partitions Z into equi ...
Solutions
... (e) (UI Freshman Math Contest, 2012) Prove that there exists no power of 2 whose decimal represenation ends in the digits 2012. (Hint: Consider congruences modulo 8.) Solution: A number ending in 2012 must be of the form n = 10000k +2012 for some integer k. Since 10000 ≡ 0 mod 8 and 2012 = 8 · 251 + ...
... (e) (UI Freshman Math Contest, 2012) Prove that there exists no power of 2 whose decimal represenation ends in the digits 2012. (Hint: Consider congruences modulo 8.) Solution: A number ending in 2012 must be of the form n = 10000k +2012 for some integer k. Since 10000 ≡ 0 mod 8 and 2012 = 8 · 251 + ...
Lecture notes, sections 2.5 to 2.7
... Let us now prove a few simple theorems, to show how congruences can be used. The important thing in these examples is not the statement of the theorem, but the method of proof, using congruences. Theorem 16. Every odd square gives the remainder 1 when divided by 4. Proof. (Recall that a square is an ...
... Let us now prove a few simple theorems, to show how congruences can be used. The important thing in these examples is not the statement of the theorem, but the method of proof, using congruences. Theorem 16. Every odd square gives the remainder 1 when divided by 4. Proof. (Recall that a square is an ...
Basic Combinatorics - Math - The University of Tennessee, Knoxville
... The function f : A → B is injective (in older terminology, one-to-one ) if a1 6= a2 ⇒ f (a1 ) 6= f (a2 ), i.e., if distinct elements of A are always mapped by f to distinct elements of B. Taking the contrapositive of the above definition, we get an occasionally useful equivalent formulation of injec ...
... The function f : A → B is injective (in older terminology, one-to-one ) if a1 6= a2 ⇒ f (a1 ) 6= f (a2 ), i.e., if distinct elements of A are always mapped by f to distinct elements of B. Taking the contrapositive of the above definition, we get an occasionally useful equivalent formulation of injec ...
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?