From hip-hop!benjie@amdahl.com Sun Apr 3 08:51:00 1994 From: David G Radcliffe Subject: Hubey's paper, FTP by email, and the Pythagorean Theorem I have read Prof. Hubey's manuscript. It would be an understatement to say that it is filled with errors; indeed it is difficult to find any correct arguments in it. The funniest statement in the paper is that there exist repeating decimals with INFINITE period -- apparently they start repeating at the infinity-plus-first place... One person recently said that he was unable to get this paper because he did not have access to ftp. But it is possible to access ftp by email. Send an email message to ftpmail@decwrl.dec.com with the word "help" in the message body to get more information. I recently discovered a utility on my system which converts Postscript files to ascii. So, I have done this with Hubey's paper. You can retrieve it by anonymous ftp from alpha2.csd.uwm.edu. The file name is pub/hubey.txt. Note that the formulas and diagrams have been butchered, and may be indecipherable. The original paper is at amiga.montclair.edu:/pub/hilbert.ps.Z . On a completely unrelated topic, I just read an amazing proof of the Pythagorean Theorem that I would like to share. -------------- From: conway@math.Princeton.EDU (John Conway) Subject: Re: POW Solution, March 21-25 By the way, this gives a nice proof of Pythagoras. Use the triangle itself: B A D C The triangles ABC, ADB, BDC are all similar (if B is the right angle of ABC). But plainly area ABC = area ABD + area BCD Now replace the triangles by squares or any other similar figures! I think some students might like to see this! JHC From hip-hop!benjie@amdahl.com Mon Apr 4 13:41:00 1994 From: efedula@aol.com (EFEDULA) Subject: '94 AIME Did anyone else here take ther AIME? How'd it go for you and your school? Everyone I've talked to from my school (except for one person) pretty much bombed it. I think our average score will be about .1. I got at least a 2, and Alex Saltman got at least a 6. Somebody else thinks they got a 3, and nobody else has claimed to have a score higher than 1 (although there could be more that I just haven't talked to). There are still three problems I need to check (if I could retake the test right now, I'd get at least a 12). If there aren't any schools still waiting to take it, perhaps we could get a discussion of the problems going... E -I'm gonna be out of town on the day of the USAMO anyway, so I intentionally missed all the easy problems to avoid having a major conflict. Yeah, that's the ticket... :) From hip-hop!benjie@amdahl.com Sun Apr 10 06:58:00 1994 From: benjie@hh.sbay.org (Benjie KE6BCU) Subject: Bulletin ******** Bulletin ******** From: naidu@waikato.ac.nz (Ram) Subject: vedic mathematics VEDIC MATHEMATICS as described by Swami Tirthaji I am a post-graduate education student at Waikato University, New Zealand. I am interested in making contact with researchers (including Critics), and posibly establishing a bulletin board, in the area of Vedic Mathematics (Indian/Hindu Maths). At the moment I am looking to improve my reading base and journal article file on the topic of vedic mathematics. Would anyone aware of articles/journals please let me know titles and sources. E-Mail Addresses for Universities in India Also if anyone has got E-Mail addresses for any Universities in India could let me know please. Please reply to; naidu@waikato.ac.nz From: Theofanis Kontogiannis Subject: Dynamical programming. Anyone knows where can I find some programms for Dynamic(al) Programming? Thank you. From hip-hop!benjie@amdahl.com Sun Apr 10 06:54:00 1994 From: benjie@hh.sbay.org (Benjie KE6BCU) Subject: AIME Problems I will post the first two AIME problems today. More to come later. 1. The increasing sequence 3, 15, 24, 48, ... consists of those positive multiples of 3 that are on less than a perfet square. What is the remainder when the 1994th term of the sequence is divided by 1000? 2. A circle with diameter PQ of length 10 is internally tangent at P to a circle of radius 20. Square ABCD is constructed with A and B on the larger circle, CD tangent at Q to the smaller circle, and the smaller circle outside ABCD. The length of AB can be written in the form m+sqrt(n), where m and n are integers. Find m+n. Have fun, Benjie From hip-hop!benjie@amdahl.com Tue Apr 12 14:58:00 1994 From: root@isaac.msfc.nasa.gov (Operator) Subject: Re: AIME Problems >1. The increasing sequence 3, 15, 24, 48, ... consists of those >positive multiples of 3 that are on less than a perfet square. What >is the remainder when the 1994th term of the sequence is divided by >1000? The remainder is 0.063; the number is 8952063. I must admit that I did this by computer, evaluating the expression (n^2-1)/3 - INT(n^2-1)/3 for n=2,3,4,... >2. A circle with diameter PQ of length 10 is internally tangent at P >to a circle of radius 20. Square ABCD is constructed with A and B >on >the larger circle, CD tangent at Q to the smaller circle, and the >smaller circle outside ABCD. The length of AB can be written in the >form m+sqrt(n), where m and n are integers. Find m+n. m=8, n=304. I first drew a diagram of the problem that indicated that AB had to be >20 and <30. Further consideration of the problem allowed me to find a right triangle with sides AB/2, x, and hypotenuse 20 (the radius of the large circle). As it turned out a rectangle could also be constructed allowing AB to be expressed as 10 + x. Expressing x in terms of AB led to a quadratic with solutions 8 +/- sqrt(304). This is my first submission to iams; I hope I'm doing this right. Hopefully someone will tell me if I'm mailing to the wrong address. Michael Flora NASA - Marshall Space Flight Center From hip-hop!benjie@amdahl.com Tue Apr 12 15:05:00 1994 From: Heiner Marxen Subject: Re: AIME Problems In article you write: >2. A circle with diameter PQ of length 10 is internally tangent at P >to a circle of radius 20. Square ABCD is constructed with A and B on >the larger circle, CD tangent at Q to the smaller circle, and the >smaller circle outside ABCD. The length of AB can be written in the >form m+sqrt(n), where m and n are integers. Find m+n. Let M be the center of the larger circle: PM = 20. Because of symmetry with respect to the line through PQ M is on that line: QM = 20-10 = 10. Let E be the point of intersection of AB with the line through PQ. Because of symmetry E halves AB: AE = AB/2. Let a = AE. As QE = AB = 2a: ME = QE - QM = 2a - 10. Now we have a triangle AEM with a right angle at E. As AM = 20 (radius of larger circle), a^2 + (2a-10)^2 = 20^2. Solving for a: a = 4 +- sqrt(76). Excluding the negative solution: a = 4 + sqrt(76). Hence AB = 2a = 8 + sqrt(304). m=8, n=304, m+n = 312. Can the sum m+n be found more easily than to find both m and n? -- Heiner Marxen heiner@drb.insel.de From hip-hop!benjie@amdahl.com Wed Apr 13 08:42:00 1994 From: "AASMUND KVAMME" Subject: Radius of inscribed circle One of my students came up with this problem: Given three circles with radius 1, 2 and 3. They are placed so that each is tangent to the two others. This leaves a small "hole" between the circles. What is the radius of the largest circle that fits in this "hole"? Aasmund Kvamme Bergen College of Engineering Lars Hilesgt. 38 N-5008 Bergen Norway e-mail: ak@mango.bih.no From hip-hop!benjie@amdahl.com Thu Apr 14 14:13:00 1994 From: peterw@cco.caltech.edu (Peter T. Wang) Subject: Re: AIME Problems (Spoiler #1) Because the AIME does not permit the usage of calculators, a less computationally intensive solution follows: Note that all positive integers can be expressed in the form 3k, 3k+1, or 3k+2 where k is some positive integer. Then x^2(mod 3) can be one of (3k)^2=9k^2==0(mod 3), (3k+1)^2=9k^2+6k+1==1(mod 3), or (3k+2)^2 =9k^2+12k+4=3(3k^2+4k+1)+1==1(mod 3). Therefore each term of the sequence 3,15,24,48,... is of the form (3k+1)^2-1 or (3k+2)^2-1. To this end, note that the first term of the sequence corresponds to k=0 in the second form; i.e. 3=(3(0)+2)^2-1. Hence the 1994th term is of the first form where k=1994/2=997. But 997=1000-3; hence x_1994 = (3(1000-3)+1)^2-1 and x_1994 (mod 1000) = (3000-8)^2-1 = 9*1000^2-48*1000+64-1 = 1000(9000-48)+63 == 63(mod 1000). Peter Wang From hip-hop!benjie@amdahl.com Thu Apr 14 14:17:00 1994 From: efedula@aol.com (EFEDULA) Subject: Re: AIME Problems (Spoiler #1) In article <2ofuf0$pcv@hip-hop.hh.sbay.org>, ua532@freenet.victoria.bc.ca (David Snook) writes: --- >1. The increasing sequence 3, 15, 24, 48, ... consists of those >positive multiples of 3 that are on less than a perfet square. What >is the remainder when the 1994th term of the sequence is divided by >1000? The formula to find the N-th term at the I-th position: N = [{int((3/2) * I) + 1}^2] - 1 N = [{int((3/2) * 1994) + 1)^2] - 1 N = [2992^2]-1 N = 8,952,063 63 = N mod 1000 --- Another way to do that last step is to notice that 2992 is congruent to -8 (mod 1000), and (-8)^2=64. Then subtracting 1 yields the correct answer of 063. E From hip-hop!benjie@amdahl.com Thu Apr 14 14:21:00 1994 From: efedula@aol.com (EFEDULA) Subject: Interesting problems Here are a few interesting problems that I've run across recently (usually while in the process of solving another problem): Find or disprove the existance of all right triangles where the lengths of all three altitudes are integers. Prove that the product of one of the sides (not bases) and a diagonal is the same for all trapezoids inscribed in the same circle. Prove that all even perfect numbers greater than 6 are congruent to 1 mod 9. Find or disprove the existance of all bricks (rectangular prism, or whatever you want to call it) where all three side lengths, all three external diagonal lengths, and the internal diagonal length are all integers (I believe this is an unsolved problem). E From hip-hop!benjie@amdahl.com Thu Apr 14 14:23:00 1994 From: gt0095a@prism.gatech.edu (Walter Sun) Subject: Posting Messages 1. The increasing sequence 3, 15, 24, 48, ... consists of those positive multiples of 3 that are one less than a perfet square. What is the remainder when the 1994th term of the sequence is divided by 1000? All numbers one less than a perfect square may be written as x^2 - 1, factorable into (x+1)(x-1). For a number like this to be divisible by 3, either x+1 or x-1 must be divisible by 3. This case occurs whenever x is not divisible by three (because when it isn't divisible by 3, it is either 1 more than a factor of 3 or one less than a factor of 3). The first couple of values of x are 2,4,5,7,8,10,11, etc. The odd number terms may be determined with the relation 3n-1 for the nth odd term. Likewise, the even number terms may be determined with the relation 3n+1 for the nth even number. 1994 is the 1994/2 = 997th even number. Thus, the value for x is 3(997) + 1 = 2992. The value of x^2 - 1 (the 1994th term in the sequence) may be written as (3000 - 8)^2 - 1 = 3000^2 - 2*8*3000 + 64 - 1. When this number is divided by 1000, the first two terms yield no remainders. Thus, the remainder is just 64 -1 = 63. Walter Sun gt0095a@prism.gatech.edu From hip-hop!benjie@amdahl.com Mon Apr 18 04:04:00 1994 From: tarrant dennis wayne/jr. Subject: Even perfect numbers =1 (mod 9) Proof of even perfect numbers =1 (mod 9) to follow. 10 9 8 7 6 5 4 3 2 1 First, note that even perfect numbers are of the form: n=2^(p-1)*(2^p-1) where 2^p-1 is a prime, and, thus, so is p. Also, note that odd primes are of the form 6k+1 or 6k+5. Case 1: p=6k+1 then, n=2^(6k)*(2^(6k+1))-1 2^6=64=1 (mod 9), so 2^6k=1 (mod 9). Also, 2^(6k+1)-1=2-1=1 (mod 9) thus, 2^(6k)*(2^(6k+1)-1)=1 (mod 9) Case 2:p=6k+5 2^(p-1)=7 (mod 9) 2^p -1=5-1=4 (mod 9) hence, n=7*4=28=1 (mod 9) Therefore, all even perfect numbers greater than 6 are congruent to 1 (mod 9). QED Wayne Tarrant "There is an astonishing imagination, even in the science of mathematics.... We repeat, there is far more imagination in the head of Archimedes than in that of Homer." -Francois Marie Arouet de Voltaire From hip-hop!benjie@amdahl.com Mon Apr 18 04:05:00 1994 From: efedula@aol.com (EFEDULA) Subject: Re: Interesting problems In article <2ol4om$is7@hip-hop.hh.sbay.org>, efedula@aol.com (EFEDULA) writes: --- Prove that the product of one of the sides (not bases) and a diagonal is the same for all trapezoids inscribed in the same circle. --- CORRECTION: that last part should be "all trapezoid inscribed in the same circle that have the same height." Since I'm already posting, I might as well go ahead and add another problem that I came across a few days ago... Somebody once conjectured that for every integer n, at least one of 6n+1 and 6n-1 is prime. Prove that there are infinitely many counterexamples to this conjecture. E From hip-hop!benjie@amdahl.com Mon Apr 18 03:59:00 1994 From: Kermit Rose Subject: mistake in summation formula in book. In the book Table of Integrals, Series, and Products, on page 1, it gives the formula for n-1 sum (k^2) (x^k) = 1 + 4 x^2 + 9 x^3 + ... + (n-1)^ x^(n-1) k=1 [ (- n^2 + 2n - 1) x^(n+2) +(2 n^2 - 2n -1) x^(n+1) - n^2 x^n + n^2 + x] = ------------------------------------------------------------------------ (1-x)^3 This formula is wrong! The puzzle for you is to determine: (1) One or more quick checks that would show you that the formula is wrong. (2) What was the typing error in the formula? Kermit. From hip-hop!benjie@amdahl.com Tue Apr 19 01:56:00 1994 From: jhardin@gandalf.pnl.gov (JW Hardin) Subject: Integer question Dear all, Here is a strange integer type question I was recently asked. I would make this a contest, but I don't know the answers! Rather, I was hoping that you all might shed some light on this one for me. I asked this same question on sci.math, but received no replies. Take an integer made up of a string of digits d1 through dn: A = d1d2d3...dn Form second integer, B, by rotating most significant digit of A to be the least significant digit. We will say that B = t(A) where t() is this rotation transformation. B = d2d3...dnd1 QUESTIONS: 1) What is the smallest integer A such that t(A) = 2A? 2) Is there a general rule for finding A given n (number of digits)? 3) How about t(A)=3A or t(A)=4A? 4) Are there no such numbers? If not, why not? Would it worked if we searched for t(t(A))=2A and limited n>=3? I should say that I tried to brute force this search on my computer and did not find an integer A in my search (though I limited it to those numbers less that about 10^12). This problem is keeping me up nights searching for a rule or something. I am still receiving sporadic e-mail from iams. I subscribed to the alt.math.iams newsgroup, but my site has yet to pick up a single post. Is the newsgroup active yet? Should I be asking my local news admin about this? From hip-hop!benjie@amdahl.com Thu Apr 21 03:11:00 1994 From: prknm15@ecx.tuwien.ac.at (numath) Subject: Re: Integer question JW Hardin (jhardin@gandalf.pnl.gov) wrote: : Dear all, : Here is a strange integer type question I was recently asked. I would : make this a contest, but I don't know the answers! Rather, I was hoping : that you all might shed some light on this one for me. I asked this : same question on sci.math, but received no replies. : Take an integer made up of a string of digits d1 through dn: : A = d1d2d3...dn : Form second integer, B, by rotating most significant digit of A to : be the least significant digit. We will say that B = t(A) where t() : is this rotation transformation. : B = d2d3...dnd1 : QUESTIONS: : 1) What is the smallest integer A such that t(A) = 2A? : 2) Is there a general rule for finding A given n (number of digits)? : 3) How about t(A)=3A or t(A)=4A? : 4) Are there no such numbers? If not, why not? Would it worked if : we searched for t(t(A))=2A and limited n>=3? in austria there is a scientific news-magazine that is distributed among teachers on schools and universities and i happen to get it regularly from one. in this magazine there is a problem-part where people can write problems or their solutions to problems already printed. and once there was an example nearly equal to yours (and i solved it for any single-digit-integer d whether there exists a n such that t(n) = d*n or n = d* t(n) and i got rather big results for the existing ones the way is to represent n as 10*a +b or 10^p*b + a where b is single-digited and a is of lenght p t(n) is then each the other one when rotating to the wanted direction then you make up an equation and use divisibilities and in one of the problems you get to a question like for which p is 10^p-1 divisible by (b*10-1) or something the like : I should say that I tried to brute force this search on my computer : and did not find an integer A in my search (though I limited it to those : numbers less that about 10^12). This problem is keeping me up nights : searching for a rule or something. In short: there is no sense in searching brute force if one is interested in the complete solution i will either follow it up or mail it to anyone who asks me From hip-hop!benjie@amdahl.com Thu Apr 21 03:12:00 1994 From: benjie@wales.sbay.org (Benjie KE6BCU) 3. The function f has the property that, for each real number x, f(x) + f(x-1) = x^2 If f(19)=94, what is the remainder when f(94) is divided by 1000? 4. Find the positive integer n for which [log_2 1]+[log_2 2]+...+[log_2 n] = 1994 where [x] is the greatest integer <= x. Enjoy, happy mathing Benjie From hip-hop!benjie@amdahl.com Sat Apr 23 08:30:00 1994 From: vhansen@ipfy.bau-verm.uni-karlsruhe.de (Wolfgang von Hansen) Subject: Multiple Mean Values? Hi all! Here's a problem I am currently trying to solve (but still failing :-( Given are N w-weighted (sp?) points P in an n-dimensional space. N (very) large; w integer, n small (2-3). Find k (k given, k << N) points Q that approximate the points P best. N This means that Sum[w * dd(N ,Q )] = min. Q is the nearest Q to P (1<=j<=k). i=1 i j j i dd(P,Q) is the sqared distance between P and Q: n dd(P,Q) = Sum[(P -Q )^2]. (Here the indices stand for the n components) i=1 i i Examples: If k = 1 then Q would be the weighted mean value of all P. 1 If k > 1 and the P are arranged in k "hyper"-clouds then Q should be the weighted mean values of the clouds. i Problems arise when the clouds are intersecting or when there are not enough points Q allowed (maybe 3 non-intersecting clouds and k = 2). Another question is: Find the minimum number k <= K which approximates the points P good (increasing k by one wouldn't give a much better result). K is given and K << N. Wolfgang -- vhansen@ipf.bau-verm.uni-karlsruhe.de | Gurus use `cat >a.out' instead of gcc float o=0.075,h=1.5,T,r,O,l,I;int _,L=80,s=3200;main(){for(;s%L|| (h-=o,T= -2),s;4 -(r=O*O)<(l=I*I)|++ _==L&&putchar(*((--s%L?_ Subject: [SPOILER: AIME94 problem 3] integer property > > > 3. The function f has the property that, for each real number x, > f(x) + f(x-1) = x^2 > If f(19)=94, what is the remainder when f(94) is divided by 1000? > f(0) = a f(1) + f(0) = 1 f(1) = 1 - a f(2) = 4 - f(1) = 3 + a f(3) = 9 - f(2) = 6 - a f(4) = 16 - f(3) = 10 + a for even k, f(k) = k(k+1)/2 + a for odd k, f(k) = k(k+1)/2 - a f(19) = 190 - a = 94 a = 190 - 94 = 96 f(94) = 94 * 95 /2 + 96 = 47 * 95 + 95 + 1 = 48 * 95 + 1 = 48 * (100 - 5) + 1 = 4800 - 240 + 1 = 1561 Remainder when divided by 1000 = 561. From hip-hop!benjie@amdahl.com Sat Apr 23 13:25:00 1994 From: benjie@hh.sbay.org (Benjie KE6BCU) Subject: no subject (file transmission) ------------------------------------------------------------------ From: wan@mth.msu.edu (Dai Wan) Subject: Looking for alumni from Dept. of Math in Jilin University ------------------------------------------------------------------ To the oversea alumni of the Department of Mathematics of Jilin University: We, the alumni of the Department of Mathematics of Jilin University working or studying at Michigan State Univeristy, propose to establish the Oversea Mathematical Alumni Association of Jilin University (OMAAJU). The OMAAJU intend to be an unofficial organization for the oversea alumni of the Department of Mathematics of Jilin University who want to keep in touch, get acquainted with each other, and exchange valuable information. We strongly believe that it is a good way to establish the alumni association to help us stay in touch with people who meant so much during your college experience, to help us keep abreast of forestry issues, to advance our professional development. As alumni, we cordially invite you to join the coming alumni association. Your support and participation are greatly appreciated in advance. As the begining step, we will set up a directory of the oversea alumni of JiDa-Math as soon as possible, if you are interested in the directory and the coming alumni association, and would like to be listed in the directory, please send your following information to tjwang@math.msu.edu and wan@math.msu.edu : (1) name. (2) current mailng and e-mail addresses, phone number(s). (3) enrolling year, speciality and degree(s) in Ji-Da. (4) your oversea institution, sepciality and research fields and degree(s). (5) current affiliation and position (optional). (6) working and research interest -- keywords (optional). If you have any comments and suggestions or questions, please call Zhengfang Zhou at (517)-353-6744(Office) or (517)-349-6801(Home), or send e-mail to zfzhou@math.msu.edu . Thank you for your attention. Please spread this message to other former alumni of JiDa-Math, especially those who are maintaining a class directory. Let us help each other by making this directory up to date. We are looking forward to hearing from you soon. Sincerely yours, Tianjun Wang Dai Wan ---------------------------- From: GAMBLE_L@HCCS.CC.TX.US Subject: Info on Goldbach ---------------------------- Help, Can anyone direct me to a source of information on the mathematician Eric Goldbach. Thanks From hip-hop!benjie@amdahl.com Sun Apr 24 10:52:00 1994 From: Ariel Scolnicov Subject: Re: Goldbach Somebody (GAMBLE_L@HCCS.CC.TX.US) was asking about Goldbach. Here's a nice theorem/puzzle, attributed to Goldbach by Graham, Knuth et al in their book _Concrete Mathematics_: Let P be the set of "perfect powers": P = {m^n | m, n >= 2}. The first few elements of P are 4, 8, 9, 16, 27, 32. Note that P is a *set*, so for instance 16=2^4=4^2 appears just once. Then --- \ / 1/(k-1) = 1 --- k in P Of course, there's also the better known Goldbach Conjecture (every even number is a sum of 2 primes). Does anybody know the status of this one? Ariel Scolnicov From hip-hop!benjie@amdahl.com Mon Apr 25 13:55:00 1994 From: mknapp@isp.nwu.edu (Mike Knapp) Subject: Re: Goldbach > > Of course, there's also the better known Goldbach Conjecture (every > even number is a sum of 2 primes). Does anybody know the status of > this one? > This is still an open problem. The latest result came (I believe) last year, when somebody proved that every even number is the sum of two integers, one of which is prime, and the other is either a prime or a product of two primes. Mike ------------------------------------------------------------------------------- | "The fight is what's important. Doing Michael P. Knapp | what is good and right is why we're here. | The end is variable and relative. A person mknapp@isp.nwu.edu | can only do what his heart tells him and mknapp@casbah.acns.nwu.edu | break away from the terrible things | around him. We do what we can ... and live | with the consequences. | -Marc Zanoni ------------------------------------------------------------------------------- From hip-hop!benjie@amdahl.com Tue Apr 26 14:03:00 1994 From: root@isaac.msfc.nasa.gov (Operator) Subject: Re: Integer question This may be a moribund question now but I have only recently been able to look at my iams mail. JWHardin posted: Take an integer made up of a string of digits d1 through dn: A = d1d2d3...dn Form second integer, B, by rotating most significant digit of A to be the least significant digit. We will say that B = t(A) where t() is this rotation transformation. B = d2d3...dnd1 QUESTIONS: 1) What is the smallest integer A such that t(A) = 2A? 2) Is there a general rule for finding A given n (number of digits)? 3) How about t(A)=3A or t(A)=4A? 4) Are there no such numbers? If not, why not? Would it worked if we searched for t(t(A))=2A and limited n>=3? I considered only the first and last digits of the number: d1....dn. Clearly, for the case t(A) = 2A, dn must be at least 2x d1 but less than 4x d1. Also, d1 must be even. Thus, we have only a very limited range of possibilities: 1st digit: 2 4 2nd digit: 4,5 8,9 The possible combinations are: 2....4, 2....5, 4....8, 4....9. Swapping the digits would in no case lead to the t(A) = 2A transformation. I don't have the time to do it right now but I believe the same holds for the t(A) = 3A and t(A) = 4A cases. Michael Flora NASA - Marshall Space Flight Center From hip-hop!benjie@amdahl.com Wed Apr 27 13:22:00 1994 From: drosner@mail.sas.upenn.edu (Daniel Rosner) Subject: Re: Goldbach I just did a report on the Goldbach conjecture, so I know a little bit about the current status of it. The theorem that every even number can be expresed as the sum of one prime and one number with at most two prime factors (the theorem is abbreviated (1,2)) was proven not last year, but in 1966 by a mathematician named Chen. The earliest theorem of this sort was (8,8) proven in 1919 by Brun. In 1921-22, Hardy and Littlewood almost proved the three primes theorem, that every odd number can be expressed as the sum of at most three primes. (This result is obviously implied by the Goldbach conjecture.) The problem with their proof was that they assumed a modified version of the Riemann hypothesis. In 1930 it was proven that ever even number can be expressed as the sum of at most c primes, for some constant c. (c does not depend on the even number. The theorem is trivial otherwise.) Similar methods have shown that c is less than or equalt to 6. In 1937 Vinogradov removed the Riemann assumption of the Hardy-Littlewood proof, and thus confirmed the three primes theorem. This also implies that every even number can be expressed as the sum of at most 4 primes. **************************************************************************** *Dan Rosner * Now I Wanna Sniff Some Glue * *drosner@mail.sas.upenn.edu * Now I Wanna Have Something To Do * ***************************** All The Kids Wanna Sniff Some Glue * * All The Kids Want Something To Do * * - the Ramones * ************************************************ From hip-hop!benjie@amdahl.com Fri Apr 29 12:09:00 1994 From: Kermit Rose Subject: Re: mistake in summation formula in book. Solution to following problem was found by David Snook. Several people found the two typing mistakes that I made. David found these and also found the mistake I intended for the puzzle. > In the book > Table of Integrals, Series, and Products, on > page 1, it gives the formula for > > > n-1 > sum (k^2) (x^k) = 1 + 4 x^2 + 9 x^3 + ... + (n-1)^ x^(n-1) > k=1 ^^^^ ^^^ > This should have been an x. I left off the square. > > [ (- n^2 + 2n - 1) x^(n+2) +(2 n^2 - 2n -1) x^(n+1) - n^2 x^n + n^2 + x] > = ------------------------------------------------------------------------ > (1-x)^3 The n^2 should have been x^2. Apparently someone typing misread the x as an n. > > This formula is wrong! > > The puzzle for you is to determine: > (1) One or more quick checks that would show you that the formula is wrong. Substitute n = 0 or 1 or 2. substitute x = 0 or 1. For x = 0, the formula must yield zero. For x = 1, both the numerator and denominator must be zero. For n = 0, the formula must yield zero. This test also suggests correction. For n = 1, the formula must yield zero. For n = 2, the formula must yield x. > (2) What was the typing error in the formula? > Typing n where x should have been typed. > Kermit. From hip-hop!benjie@amdahl.com Tue May 3 14:22:00 1994 From: zergerm@cc4.adams.edu Subject: DIFF OF TWO CUBES A positive integer can be represented as the difference of two squares if and only if it is not congruent to 2 modulo 4. Does anyone know if the positive integers which can be represented as the difference of two CUBES have been so conveniently characterized? Monte J. Zerger mzerger@cc4.adams.edu (719) 589-7546 Adams State College Alamosa, CO 81102 From hip-hop!benjie@amdahl.com Thu May 5 14:06:00 1994 From: weix@netcom.com (Patrick Weix) Subject: Good books--Recommendations sought I have no clever puzzle today, but I do have a request: Could you all recommend some good books? I would like to read up on some areas where I feel a little deficient, such as Number Theory and Advanced Probability and Statistics. Topology would also be good. I would like books that deliver info in the middle range. I have an undergraduate degree in Mechanical Engineering (took calculus in high school and diffy q in college), and I am currently working on a combined M.D./Ph.D. in genetics and development, so somewhere between the "numbers are our friends" approach and the "pages and pages of equations" approach would be ideal. Good college textbooks are fine. Are there books that you remember reading with enjoyment? TIA, Patrick -- weix@netcom.com From hip-hop!benjie@amdahl.com Fri May 6 11:16:00 1994 From: efedula@aol.com (EFEDULA) Subject: Re: DIFF OF TWO CUBES In article <2q77uv$ikb@hip-hop.hh.sbay.org>, zergerm@cc4.adams.edu writes: > A positive integer can be represented as the difference of two squares > if and only if it is not congruent to 2 modulo 4. > Does anyone know if the positive integers which can be represented as > the difference of two CUBES have been so conveniently characterized? Well, the reason the difference of two squares cannot be 2mod4 is that all squares are 0 or 1 mod 4. Since every odd number is the difference between two consecutive squares, it's not that hard to see that the theorem is true for odd numbers (1 and 3 mod 4), and all that remains is to prove that every multiple of 4 is the difference of two squares, which is obviously true since the sum of two consecutive odd integers is a multiple of 4. Since every cube is congruent to -1, 0, or 1 mod 9, the difference of two cubes can only be 0, 1, -1, 2, or -2, so the difference of two cubes cannot be 3, 4, 5, or 6 modulo 9. As for whether or not every number congruent to -2, -1, 0, 1, or 2 mod 9 can be expressed as the difference of two cubes, I don't know (I've never thought about this before). If you try a bunch of examples at random and you don't run across any counterexamples, then you can probably assume that it's true. If you find any counterexamples, then I would guess that there isn't such a convenient characterization. If it is true, the easiest proof would probably come either from looking at the differences between consecutive cubes or by using the fact that a^3-b^3=(a-b)(a^2+ab+b^2). Again, I haven't spent much time thinking about this, so I don't know for sure whether or not this characterization works. E E From hip-hop!benjie@amdahl.com Fri May 6 11:14:00 1994 From: tmataign@hvlpa.att.com (Thierry Mataigne (MAT) +32 2 556 7323) Hi iams, The following problem is not new. It appeared several years ago in an international competition, but I have lost any record about it, and its solution: f is a function defined on N, with values in N. ( N is the natural number set: N = { 1,2,3,4,..... }; f is defined for all such positive numbers). Prove that: if for all n, f(f(n)) < f(n+1) then for all n, f(n) = n. Have fun. Thierry Mataigne From hip-hop!benjie@amdahl.com Thu May 12 15:26:00 1994 From: benjie@hh.sbay.org (Benjie KE6BCU) Subject: Volume of a polyhedral (sp?) Given a perfect polyhedral, that is, the surface area of each surface is the same, dimension of one of its surface and total number of surfaces, how do you find out the volume of the polyhedral? (I dunnt know if it's really called polyhedral, but it suppose to be a three dimensional object with n>4 number of faces.) Benjie From hip-hop!benjie@amdahl.com Fri May 20 13:02:00 1994 From: "David A. Wagner" Subject: A quick question or two Here's two random quick questions to tickle your brain as summer approaches. The first one is from a final in analysis I took today. It had two parts, and was true or false. (a) True or False: There exists a (Lebesgue) integrable function f on the real line so that for any continuous function g, \int |f-g| > 0. (b) True or False: There exists a (Lebesgue) integrable function f on the real line and a real number e>0 so that for any continuous function g, \int |f-g| > e. The second question is one that has been lingering in my mind about Galois theory. Let z_n be a primitive n-th root of unity. I can prove that the field extension Q(z_n):Q has Galois group (Z/nZ)* for all prime n. Is it true for any integer n? If so, how do you prove it? [Here Q denotes the rational numbers and Q(z_n) stands for the rational numbers with the n-th roots of unity adjoined. (Z/nZ)* is the multiplicative group of the integers modulo n.] ------------------------------------------------------------------------------- David Wagner dawagner@princeton.edu