tag:blogger.com,1999:blog-300411792017-07-20T12:33:20.608+05:30Fun with math<b>Simple math problems for recreation and research</b><br>
"And yet the relation appears,
A small relation expanding like the shade
of a cloud on the sand,a shape on side of the hill"
-Wallace Stevens,'Connoisseur of Chaos'<br>You shall find a [S] at the end of a post if solution can be found at one of its comments.Feel free to comment.<br>
<a href="http://fun-with-math.blogspot.com/">HOME</a>Gangotryihttp://www.blogger.com/profile/03987426697873459849noreply@blogger.comBlogger46125tag:blogger.com,1999:blog-30041179.post-68310178460425058312007-12-03T15:21:00.000+05:302007-12-03T15:37:49.822+05:30Goodbye math blogTo whomsoever it may concern,I have finally decided to stop writing this math blog. This has no bearing on rest of the contributors. It has been a great journey, I see myself today with a lot more mathematical maturity as I go through old to new posts. The reason to stop is not because I am getting absorbed into more serious math but the lack of purpose for which the blog was created nearly two talegari (ತಾಳೆಗರಿ)noreply@blogger.com53tag:blogger.com,1999:blog-30041179.post-64422945570971195472007-11-22T13:51:00.000+05:302007-11-22T15:31:13.954+05:30New books in mathematicsYou can find a lot of free e-books here.Books range from math classics to recently released ones gracefully violating copyrights.I simply believe knowledge should be free.Some good books of my interest:1.The art of Counting - Paul Erdos.2.Combinatorics - Peter Cameron.3.A course in combinatorics - van Lint and Wilson4.Enumerative Combinatorics - R P Stanley5.Research Problems in Discrete geometrytalegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-60889479554488790782007-11-20T15:41:00.000+05:302007-11-20T16:32:27.437+05:30klien's problemI found this problem asing to be attacked by PHP, later I learnt this was generalized by Erdos and Szekeres(happy ending problem). Let there be 5 points arbitrarily lying on a plane.Prove that we can find four points that form a convex quadrilateral.classic proof: If the smallest convex polygon encompassing all the points is quadrilateral or pentagon, then we are done. Else, if it is a triangle (talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-21102529786961195902007-11-19T11:09:00.000+05:302007-11-19T11:29:13.724+05:30Hit by PHP (Pigeon hole principle)Problem: For coprime a and b, a/b has period of reccurance of length atmost b-1.This next one is a classic.Let there be n integers not necessarily distinct.Prove that we pick k (< (n+1)) integers which is divisible by n. Let i_1, i_2,..., i_n be the integers.Let a_1=i_1, a_2=i_1 + i_2 . . a_n=i_1+..+i_nIf one of the a_j ( j=1,..,n ) is divisible by n we are done,else by php we can talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-31072019282976511102007-11-14T10:01:00.000+05:302007-11-14T10:15:45.376+05:30Yet another counting problemThis seems to fall for computation, I am looking at a possible elegant solution.Problem:Let us have a n*n square.Two squares(both of dimension less than n*n)form an unordered pair if (i) They have atleast one 1*1 square common.(ii)One is not contained in the other.Compute the number of unordered pairs.(nothing sacred about squares, rectangles must make the problem tougher)talegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-807733418878422582007-11-02T12:34:00.000+05:302007-11-02T17:29:26.625+05:30Computing a few Invertible matricesProblem: Enumerate and generate the number of invertible matrices (n*n) with each entry either 0 or 1.Example: For n=2, we have a set of 6 matrices. They areInterestingly, these matrices form a group under matrix multiplication.A much more harder problem would be to enumerate and generate all invertible (n*n) matrices with enrites fron the set { 0,1,...,n-1 }.talegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-48230807444740223642007-10-30T19:55:00.000+05:302007-11-12T17:05:17.464+05:30Linear IndependenceThis problem is the generalization of exercise 2.3.9 from 'linear algebra' Hoffman and Kunze.Let the set of vectors be linearly independent. Prove that are also linearly independent givenI meanand so on.ps: I proved it by proving a certain matrix to be invertible and that was pretty cumbersome. Hoping for a better pretty solution :)talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-59677407658157981572007-10-23T13:17:00.000+05:302007-10-23T14:02:03.320+05:30A closed form for nth Fibonacci numberI have made an attempt to arrive at a closed form by setting up a bijection in a counting process.Consider the set . In how many ways can we select non null subset of such that no two elements in the subset give difference one ?Let be the number of ways of selecting such a subset. Depending upon whether is included in the subset or not , we getwith and It is clear that = (n+1)th Fibonaccitalegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-70040327156721752602007-10-23T13:11:00.000+05:302007-10-23T13:14:12.149+05:30Oh ! I forgot to tellI would like to welcome our new Contributor - Gangotryi.talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-84124817090310449692007-10-23T13:02:00.000+05:302007-10-23T13:03:33.555+05:30The Vodafone problemI was lazily lying on my bed thinking soon after getting up still drowsy...I have change my hutch[no vodafone] account from postpaid to prepaid. Then this thing flashed to me:1. Now vodafone announces a scheme by which you may choose number/s and you can make calls to that number/s at concessional rate. Lets call this condition as you two are 'connected'.You choosing that number/s also means talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-41420382343680797442007-10-10T15:14:00.000+05:302007-10-10T16:15:02.849+05:30An intriguing one -IIThis is the continuation of the post 'An intriguing one'. Here I make an attempt to look at a possible approach.Consider two concentric circles with radii and . It is clear that there cannot exist two points within the b-circle.I go on placing points on the circumference of a-circle with gap of at least 'b' of course, I do not know what to do with . As of now we shall distribute evenly. This talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-83713901741830230402007-10-07T14:08:00.000+05:302007-10-08T14:01:01.443+05:30Pleasure of LatexMany many thanks to Wolverine for latex implementation script on Firefox.Link to the site.good naa.. :)ps: But I found that same cant be used on be comments page. Anybody has a idea?talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-72366455833060692982007-10-07T13:27:00.000+05:302007-10-09T11:53:42.004+05:30Distribution with restrictions [S]When I wrote this post I made a typo mistake. Here is what I intended.*******************************************************************Problem: Enumerate the number of solutions of the equationsuch that for every Lets suppose so that at least one solution is assured.*************************The flipside of this is simple. How many solutions with each ? Easy.. put balls in each of the boxestalegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-35042936517842054342007-09-10T15:02:00.000+05:302007-09-10T15:23:01.488+05:30This is not going to be simple..Consider this restricted lattice bounded by [0,0];[x,0];[x,y];[y,0]. One can move from any point to another by a path , sequence of adjacent lattice points ( like one in the bad diagram I have written). Enumerate the number of paths from [0,0] to [x,y] ( obviously within the restricted grid). What about the paths that have loops? By a loop I mean the path coming back to the point that has alreadytalegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-50159122111302626562007-09-06T17:53:00.000+05:302007-09-06T18:13:36.954+05:30Recent problemsOf late, I have been studying a lot of combinatorics and attacking problems with enthusiasm.Here are a few worth mentioning ones.1. Supposing there exists a convex polygon such that no three diagonals meet at a point, (i)find the number of intersections of the diagonals. (ii)Find the number of interior areas formed.2. Enumerate the number of ways of arranging n black and n white balls around a talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-64588040797132401262007-08-19T13:39:00.000+05:302007-08-19T13:52:18.184+05:30never resist the temptation...I am tempted to post this problem.Let S={ 1,2,...,n}A and B are non empty subsets of S.C is a subset of S with elements of form a+b where a belongs to A and b belongs to B.Give an algorithm to generate all possible set D={ A,B}talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-65548459549561232052007-06-26T16:10:00.000+05:302007-06-26T16:17:33.789+05:30Two not so easy onesTake this first one from USAMO 2004.1.Suppose are integers whose greatest common divisor is 1. Let be a set of integers with the following properties:(a) For , .(b) For (not necessarily distinct), .(c) For any integers , if , then .Prove that must be equal to the set of all integers.*****************************************************If you hit the above one, try this, as of now I haven't talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-36844856815470952262007-06-15T16:25:00.000+05:302007-06-26T16:23:27.307+05:30pretty goodProve that among any 10 consecutive integers at least one is relatively prime to each of the others.hint: picture has nothing to do with the problemtalegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-21183058546575434452007-06-12T16:02:00.000+05:302007-11-12T17:07:32.236+05:30A generalizationProve that any positive integer x lying between ak and ak+1 can be represented uniquely as followsx = ak + bk-1 ak-1 +...+bk-r ak-r +..+b0 a0where a is a positive integer >1 and each bi (non negative integer) is less than a.I generalized this based on the problems found in Niven's number theory book( pg 19 pro, 44 and 45 , fifth edition)talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-72046599977469564542007-06-07T11:38:00.000+05:302007-06-26T16:24:15.358+05:30A simple one [S]This objective type question is said to have appeared in one of the old ISI selection papers.A club with 'x' members is organized into 4 committees such that[a] each member is in exactly two committees[b] any two committees have exactly one member in commonThen,1. exactly two values both between 4 and 8.2. exactly one value lying between 4 and 8.3. exactly two values between 8 and 16.4. exactlytalegari (ತಾಳೆಗರಿ)noreply@blogger.com1tag:blogger.com,1999:blog-30041179.post-15119245282603756062007-06-05T16:06:00.000+05:302007-06-07T12:35:23.462+05:30A functional equationProve that this equation does not admit any real continuous function R ->R ( try Rn->Rm)f(x).f(x+1) + f(x) + 1 = 0My progress so far: turn this into a recursive equation f(x+1)= -1/ [ f(x) +1]suppose if f(xo)=kthen , f(xo+4)=k for arbitrary xo and k belonging to R.Let us select x and x+$ , two real numbers, thenepsilon=f(x)-f(x+$) = $/ (1+x)2//neglecting $ as it does not matter in talegari (ತಾಳೆಗರಿ)noreply@blogger.com2tag:blogger.com,1999:blog-30041179.post-53373261586176270692007-05-14T10:57:00.000+05:302007-05-14T11:01:36.651+05:30This could be toughI am told that this problem can be solved just by consideration of parity.Let g(x) be a bijective function N->N and 'k' be any positive odd integer. Prove that there does not exist any function f(x):N->N such that f (f(x)) = g(x) + ktalegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-28653708218253003692007-05-04T10:14:00.000+05:302007-05-04T10:28:08.488+05:30Some old wine...Well, take this old probability problem.A mathematician carries two matchboxes(in order to complicate life!), both initially containing 'n' sticks each.Whenever a stick is needed, he takes it from one of the boxes(boxes are distinguishable)but never keeps count of matches.Once it so happens that he takes out one of the box and finds it empty, takes out the other and that is also empty.What is thetalegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-76647081548218219712007-04-30T12:26:00.000+05:302007-09-06T18:15:04.671+05:30IISC Mathematics Integrated PhD paper 2007You may download a copy of question paper of IISC Mathematics Integrated PhD paper 2007 held on 29th apr 2007 here ( link no longer exists!)talegari (ತಾಳೆಗರಿ)noreply@blogger.com0tag:blogger.com,1999:blog-30041179.post-58241322359352935452007-04-24T19:52:00.000+05:302007-04-24T20:06:41.488+05:30Finally, an easy one [S]I caught this good one while I was having a leisurely look at mayhem taunt problems (M74).May look daunting at the first but you should be able to hit it within 10 minutes.Problem: A circle has 12 equally spaced points on its circumference. How many ways can the numbers 1 through 12 be assigned to the points so that if the points 1 through 12 are connected with line segments, in order, the talegari (ತಾಳೆಗರಿ)noreply@blogger.com1