[Math.NT] 6 May 2001 Number Neeti H Imn Yohss Eaddb Aya H M the As Many by Regarded Hypothesis, Riemann the in Interest ABSTRACT
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2020 Chapter Competition Solutions
2020 Chapter Competition Solutions Are you wondering how we could have possibly thought that a Mathlete® would be able to answer a particular Sprint Round problem without a calculator? Are you wondering how we could have possibly thought that a Mathlete would be able to answer a particular Target Round problem in less 3 minutes? Are you wondering how we could have possibly thought that a particular Team Round problem would be solved by a team of only four Mathletes? The following pages provide solutions to the Sprint, Target and Team Rounds of the 2020 MATHCOUNTS® Chapter Competition. These solutions provide creative and concise ways of solving the problems from the competition. There are certainly numerous other solutions that also lead to the correct answer, some even more creative and more concise! We encourage you to find a variety of approaches to solving these fun and challenging MATHCOUNTS problems. Special thanks to solutions author Howard Ludwig for graciously and voluntarily sharing his solutions with the MATHCOUNTS community. 2020 Chapter Competition Sprint Round 1. Each hour has 60 minutes. Therefore, 4.5 hours has 4.5 × 60 minutes = 45 × 6 minutes = 270 minutes. 2. The ratio of apples to oranges is 5 to 3. Therefore, = , where a represents the number of apples. So, 5 3 = 5 × 9 = 45 = = . Thus, apples. 3 9 45 3 3. Substituting for x and y yields 12 = 12 × × 6 = 6 × 6 = . 1 4. The minimum Category 4 speed is 130 mi/h2. The maximum Category 1 speed is 95 mi/h. Absolute difference is |130 95| mi/h = 35 mi/h. -
PRIME-PERFECT NUMBERS Paul Pollack [email protected] Carl
PRIME-PERFECT NUMBERS Paul Pollack Department of Mathematics, University of Illinois, Urbana, Illinois 61801, USA [email protected] Carl Pomerance Department of Mathematics, Dartmouth College, Hanover, New Hampshire 03755, USA [email protected] In memory of John Lewis Selfridge Abstract We discuss a relative of the perfect numbers for which it is possible to prove that there are infinitely many examples. Call a natural number n prime-perfect if n and σ(n) share the same set of distinct prime divisors. For example, all even perfect numbers are prime-perfect. We show that the count Nσ(x) of prime-perfect numbers in [1; x] satisfies estimates of the form c= log log log x 1 +o(1) exp((log x) ) ≤ Nσ(x) ≤ x 3 ; as x ! 1. We also discuss the analogous problem for the Euler function. Letting N'(x) denote the number of n ≤ x for which n and '(n) share the same set of prime factors, we show that as x ! 1, x1=2 x7=20 ≤ N (x) ≤ ; where L(x) = xlog log log x= log log x: ' L(x)1=4+o(1) We conclude by discussing some related problems posed by Harborth and Cohen. 1. Introduction P Let σ(n) := djn d be the sum of the proper divisors of n. A natural number n is called perfect if σ(n) = 2n and, more generally, multiply perfect if n j σ(n). The study of such numbers has an ancient pedigree (surveyed, e.g., in [5, Chapter 1] and [28, Chapter 1]), but many of the most interesting problems remain unsolved. -
Newsletter No. 31 March 2004
Newsletter No. 31 March 2004 Welcome to the second newsletter of 2004. We have lots of goodies for you this month so let's get under way. Maths is commonly said to be useful. The variety of its uses is wide but how many times as teachers have we heard students exclaim, "What use will this be when I leave school?" I guess it's all a matter of perspective. A teacher might say mathematics is useful because it provides him/her with a livelihood. A scientist would probably say it's the language of science and an engineer might use it for calculations necessary to build bridges. What about the rest of us? A number of surveys have shown that the majority of us only need to handle whole numbers in counting, simple addition and subtraction and decimals as they relate to money and domestic measurement. We are adept in avoiding arithmetic - calculators in their various forms can handle that. We prefer to accept so-called ball-park figures rather than make useful estimates in day-to-day dealings and computer software combined with trial-and-error takes care of any design skills we might need. At the same time we know that in our technological world numeracy and computer literacy are vital. Research mathematicians can push boundaries into the esoteric, some of it will be found useful, but we can't leave mathematical expertise to a smaller and smaller proportion of the population, no matter how much our students complain. Approaching mathematics through problem solving - real and abstract - is the philosophy of the nzmaths website. -
Formal Logic Lecture 2
Faculty of Philosophy Formal Logic Lecture 2 Peter Smith Peter Smith: Formal Logic, Lecture 2 1 Outline Validity again Systematicity and formality Modality and the invalidity principle The counterexample method Peter Smith: Formal Logic, Lecture 2 2 Valid deductions as absolutely watertight inferences I Consider this inference Every good philosopher knows some logic. No post-modernist knows any logic. Jacques is a post-modernist. So: Jacques isn't a good philosopher. Who knows whether the premisses are true or false? But if the premisses are true, the conclusion has to be true too. I The inferential move in this argument from premisses to conclusion is absolutely watertight. There is not even a remote chance, there's no possibility at all, that the premisses are true and the conclusion false. I Such an inference, we said, is deductively valid. I It is conventional to call a (one-step) argument valid if its inferential move is valid. An argument which is valid and has true premisses is called sound. Validity again Logicians' deductions vs Sherlock's deductions I The detective gathers evidence, and `deduces' who did the dastardly deed. I But this (normally) is not a valid deduction in the logician's sense. The detective's account of the murder may fit the facts and strike us as obviously the best explanation. But the best (= most plausible, most likely) explanation doesn't have to be the only one that is logically consistent with the facts. I So the `deduction' from the evidential facts to the detective's proposed explanation isn't absolutely guaranteed to be truth-preserving, i.e. -
POWERFUL AMICABLE NUMBERS 1. Introduction Let S(N) := ∑ D Be the Sum of the Proper Divisors of the Natural Number N. Two Disti
POWERFUL AMICABLE NUMBERS PAUL POLLACK P Abstract. Let s(n) := djn; d<n d denote the sum of the proper di- visors of the natural number n. Two distinct positive integers n and m are said to form an amicable pair if s(n) = m and s(m) = n; in this case, both n and m are called amicable numbers. The first example of an amicable pair, known already to the ancients, is f220; 284g. We do not know if there are infinitely many amicable pairs. In the opposite direction, Erd}osshowed in 1955 that the set of amicable numbers has asymptotic density zero. Let ` ≥ 1. A natural number n is said to be `-full (or `-powerful) if p` divides n whenever the prime p divides n. As shown by Erd}osand 1=` Szekeres in 1935, the number of `-full n ≤ x is asymptotically c`x , as x ! 1. Here c` is a positive constant depending on `. We show that for each fixed `, the set of amicable `-full numbers has relative density zero within the set of `-full numbers. 1. Introduction P Let s(n) := djn; d<n d be the sum of the proper divisors of the natural number n. Two distinct natural numbers n and m are said to form an ami- cable pair if s(n) = m and s(m) = n; in this case, both n and m are called amicable numbers. The first amicable pair, 220 and 284, was known already to the Pythagorean school. Despite their long history, we still know very little about such pairs. -
Arxiv:2106.08994V2 [Math.GM] 1 Aug 2021 Efc Ubr N30b.H Rvdta F2 If That and Proved Properties He Studied BC
Measuring Abundance with Abundancy Index Kalpok Guha∗ Presidency University, Kolkata Sourangshu Ghosh† Indian Institute of Technology Kharagpur, India Abstract A positive integer n is called perfect if σ(n) = 2n, where σ(n) denote n σ(n) the sum of divisors of . In this paper we study the ratio n . We de- I → I n σ(n) fine the function Abundancy Index : N Q with ( ) = n . Then we study different properties of Abundancy Index and discuss the set of Abundancy Index. Using this function we define a new class of num- bers known as superabundant numbers. Finally we study superabundant numbers and their connection with Riemann Hypothesis. 1 Introduction Definition 1.1. A positive integer n is called perfect if σ(n)=2n, where σ(n) denote the sum of divisors of n. The first few perfect numbers are 6, 28, 496, 8128, ... (OEIS A000396), This is a well studied topic in number theory. Euclid studied properties and nature of perfect numbers in 300 BC. He proved that if 2p −1 is a prime, then 2p−1(2p −1) is an even perfect number(Elements, Prop. IX.36). Later mathematicians have arXiv:2106.08994v2 [math.GM] 1 Aug 2021 spent years to study the properties of perfect numbers. But still many questions about perfect numbers remain unsolved. Two famous conjectures related to perfect numbers are 1. There exist infinitely many perfect numbers. Euler [1] proved that a num- ber is an even perfect numbers iff it can be written as 2p−1(2p − 1) and 2p − 1 is also a prime number. -
Arxiv:2008.10398V1 [Math.NT] 24 Aug 2020 Children He Has
JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 00, Number 0, Pages 000{000 S 0894-0347(XX)0000-0 RECURSIVELY ABUNDANT AND RECURSIVELY PERFECT NUMBERS THOMAS FINK London Institute for Mathematical Sciences, 35a South St, London W1K 2XF, UK Centre National de la Recherche Scientifique, Paris, France The divisor function σ(n) sums the divisors of n. We call n abundant when σ(n) − n > n and perfect when σ(n) − n = n. I recently introduced the recursive divisor function a(n), the recursive analog of the divisor function. It measures the extent to which a number is highly divisible into parts, such that the parts are highly divisible into subparts, so on. Just as the divisor function motivates the abundant and perfect numbers, the recursive divisor function motivates their recursive analogs, which I introduce here. A number is recursively abundant, or ample, if a(n) > n and recursively perfect, or pristine, if a(n) = n. There are striking parallels between abundant and perfect numbers and their recursive counterparts. The product of two ample numbers is ample, and ample numbers are either abundant or odd perfect numbers. Odd ample numbers exist but are rare, and I conjecture that there are such numbers not divisible by the first k primes|which is known to be true for the abundant numbers. There are infinitely many pristine numbers, but that they cannot be odd, apart from 1. Pristine numbers are the product of a power of two and odd prime solutions to certain Diophantine equations, reminiscent of how perfect numbers are the product of a power of two and a Mersenne prime. -
Proofs and Mathematical Reasoning
Proofs and Mathematical Reasoning University of Birmingham Author: Supervisors: Agata Stefanowicz Joe Kyle Michael Grove September 2014 c University of Birmingham 2014 Contents 1 Introduction 6 2 Mathematical language and symbols 6 2.1 Mathematics is a language . .6 2.2 Greek alphabet . .6 2.3 Symbols . .6 2.4 Words in mathematics . .7 3 What is a proof? 9 3.1 Writer versus reader . .9 3.2 Methods of proofs . .9 3.3 Implications and if and only if statements . 10 4 Direct proof 11 4.1 Description of method . 11 4.2 Hard parts? . 11 4.3 Examples . 11 4.4 Fallacious \proofs" . 15 4.5 Counterexamples . 16 5 Proof by cases 17 5.1 Method . 17 5.2 Hard parts? . 17 5.3 Examples of proof by cases . 17 6 Mathematical Induction 19 6.1 Method . 19 6.2 Versions of induction. 19 6.3 Hard parts? . 20 6.4 Examples of mathematical induction . 20 7 Contradiction 26 7.1 Method . 26 7.2 Hard parts? . 26 7.3 Examples of proof by contradiction . 26 8 Contrapositive 29 8.1 Method . 29 8.2 Hard parts? . 29 8.3 Examples . 29 9 Tips 31 9.1 What common mistakes do students make when trying to present the proofs? . 31 9.2 What are the reasons for mistakes? . 32 9.3 Advice to students for writing good proofs . 32 9.4 Friendly reminder . 32 c University of Birmingham 2014 10 Sets 34 10.1 Basics . 34 10.2 Subsets and power sets . 34 10.3 Cardinality and equality . -
Amicable Numbers
Amicable numbers Saud Dhaafi Mathematics Department, Western Oregon University May 11, 2021 1 / 36 • 1 is the first deficient number. • 6 is the first perfect number. • 12 is the first abundant number. • An amicable numbers are two different numbers m and n where σ(m)=σ(n) = m + n and m < n. • σ(n) is the sum of all positive divisors of a positive integer n including 1 and n itself. • A proper divisor is the function s(n) = σ(n) − n. Introduction • We know that all numbers are interesting. Zero is no amount while it is an even number. 2 / 36 • 6 is the first perfect number. • 12 is the first abundant number. • An amicable numbers are two different numbers m and n where σ(m)=σ(n) = m + n and m < n. • σ(n) is the sum of all positive divisors of a positive integer n including 1 and n itself. • A proper divisor is the function s(n) = σ(n) − n. Introduction • We know that all numbers are interesting. Zero is no amount while it is an even number. • 1 is the first deficient number. 3 / 36 • 12 is the first abundant number. • An amicable numbers are two different numbers m and n where σ(m)=σ(n) = m + n and m < n. • σ(n) is the sum of all positive divisors of a positive integer n including 1 and n itself. • A proper divisor is the function s(n) = σ(n) − n. Introduction • We know that all numbers are interesting. Zero is no amount while it is an even number. -
Groups of Piecewise Projective Homeomorphisms
Groups of piecewise projective homeomorphisms Nicolas Monod1 Section de Mathématiques, École Polytechnique Fédérale de Lausanne, 1015 Lausanne, Switzerland Edited by Gregory A. Margulis, Yale University, New Haven, CT, and approved February 5, 2013 (received for review October 22, 2012) The group of piecewise projective homeomorphisms of the line The main result of this article is the following, for which we provides straightforward torsion-free counterexamples to the so- introduce a method for proving amenability. called von Neumann conjecture. The examples are so simple that many additional properties can be established. Theorem 1. The group H(A) is nonamenable if A ≠ Z. The next result is a sequacious generalization of the corre- free groups | paradoxical decomposition | von Neumann problem sponding theorem of Brin–Squier (16) about piecewise affine transformations, and we claim no originality. n 1924, Banach and Tarski (1) accomplished a rather para- Idoxical feat. They proved that a solid ball can be decomposed Theorem 2. The group H does not contain any nonabelian free sub- fi group. Thus, H(A) inherits this property for any subring A < R. into ve pieces, which are then moved around and reassembled = R in such a way as to obtain two balls identical to the original one Thus, already H H( ) itself is a counterexample to the von Neumann conjecture. Writing H(A) as the directed union of its (1). This wellnigh miraculous duplication was based on Haus- fi dorff’s (2) 1914 work. nitely generated subgroups, we deduce Corollary 3. In his 1929 study of the Hausdorff–Banach–Tarski paradox, Corollary 3. -
Perfect, Abundant, and Deficient Numbers the Abundance Quotient Is the Sum of Its Factors (Except for the Number Itself) Divided by the Number Itself
Perfect, Abundant, and Deficient Numbers The abundance quotient is the sum of its factors (except for the number itself) divided by the number itself. For example, with 24, the sum of its factors is 36, so the abundance quotient for 24 is 36÷24 = 1.5. All perfect numbers, by definition, have an abundance quotient exactly equal to one. The first nine perfect numbers are: 6 8,589,869,056 28 137,438,691,328 496 2,305,843,008,139,952,128 8128 2,658,455,991,569,831,744,654,692,615,953,842,176. 33,550,336 The tenth perfect number has 54 digits! It is still unknown if any odd perfect number exists. Interestingly, the first 231 abundant numbers are all even numbers. The first odd-numbered abundant number is 945 (quotient = 1.032), and the second one is 1575 (quotient = 1.047). The abundance quotients of each of the “biggest” abundant numbers (i.e., having an abundance quotient greater than any previous number) from 6 up to 30,000 are listed below. 12 is abundant with a quotient of 1.333 720 is abundant with a quotient of 2.358 24 is abundant with a quotient of 1.500 840 is abundant with a quotient of 2.429 36 is abundant with a quotient of 1.528 1260 is abundant with a quotient of 2.467 48 is abundant with a quotient of 1.583 1680 is abundant with a quotient of 2.543 60 is abundant with a quotient of 1.800 2520 is abundant with a quotient of 2.714 120 is abundant with a quotient of 2.000 5040 is abundant with a quotient of 2.838 180 is abundant with a quotient of 2.033 10080 is abundant with a quotient of 2.900 240 is abundant with a quotient -
The Mertens Conjecture
The Mertens conjecture Herman J.J. te Riele∗ Title of c Higher Education Press This book April 8, 2015 and International Press ALM ??, pp. ?–? Beijing-Boston Abstract The Mertens conjecture states that |M(x)|x−1/2 < 1 for x > 1, where M(x) = P1≤n≤x µ(n) and where µ(n) is the M¨obius function defined by: µ(1) = 1, µ(n) = (−1)k if n is the product of k distinct primes, and µ(n)=0 if n is divisible by a square > 1. The truth of the Mertens conjecture implies the truth of the Riemann hypothesis and the simplicity of the complex zeros of the Riemann zeta function. This paper gives a concise survey of the history and state-of-affairs concerning the Mertens conjecture. Serious doubts concerning this conjecture were raised by Ingham in 1942 [12]. In 1985, the Mertens conjecture was disproved by Odlyzko and Te Riele [23] by making use of the lattice basis reduction algorithm of Lenstra, Lenstra and Lov´asz [19]. The best known results today are that |M(x)|x−1/2 ≥ 1.6383 and there exists an x< exp(1.004 × 1033) for which |M(x)|x−1/2 > 1.0088. 2000 Mathematics Subject Classification: Primary 11-04, 11A15, 11M26, 11Y11, 11Y35 Keywords and Phrases: Mertens conjecture, Riemann hypothesis, zeros of the Riemann zeta function 1 Introduction Let µ(n) be the M¨obius function with values µ(1) = 1, µ(n) = ( 1)k if n is the product of k distinct primes, and µ(n) = 0 if n is divisible by a square− > 1.