<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearch/1.1/' xmlns:georss='http://www.georss.org/georss' xmlns:thr='http://purl.org/syndication/thread/1.0' xmlns:gd='http://schemas.google.com/g/2005' gd:etag='W/&quot;D0MFRXc8eyp7ImA9Wx9VEEw.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386</id><updated>2011-01-25T20:16:54.973-08:00</updated><title>Equation Solver</title><subtitle type='html'></subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://www.equationsolver.info/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default?redirect=false&amp;v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><link rel='next' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default?start-index=26&amp;max-results=25&amp;redirect=false&amp;v=2'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>628</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>25</openSearch:itemsPerPage><entry gd:etag='W/&quot;DEEBRXg_eSp7ImA9Wx9QEUo.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-5451175253429491864</id><published>2010-12-19T17:17:00.000-08:00</published><updated>2010-12-23T23:44:14.641-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-23T23:44:14.641-08:00</app:edited><title>How to do double digit long division step by step?</title><summary>How to do double digit long division step by step?

More books about long division for kids



How to do Long Division with Remainders?
When we are given a long division to do it will not always work out to a         whole number. Sometimes there will be numbers left over. These are known         as remainders. Taking an example similar to that on the Long         Division page it becomes more </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5451175253429491864?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5451175253429491864?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-double-digit-long-division.html' title='How to do double digit long division step by step?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_J56MO7usZSU/TRRK9n_V8tI/AAAAAAAAAvk/BqGI19afso8/s72-c/divide15.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQX04cCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1015160908598198114</id><published>2010-12-17T02:33:00.000-08:00</published><updated>2010-12-21T02:10:50.338-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.338-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Multiplication'/><title>How to do long multiplication with decimals?</title><summary>Decimal values are used to symbolize fraction numbers. In the  multiplication with decimals are writing exclusive of a fraction having  denominator and denominator. Decimal values are might be greater than 1  or less than 1. For the model, the fraction 4/10 might be written as the  decimal value as 0.4 and marked as four tenths or zero point four. The  decimal of 45/10 is 4.5 which can be marked </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1015160908598198114?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1015160908598198114?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-multiplication-with.html' title='How to do long multiplication with decimals?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry><entry gd:etag='W/&quot;DEUBQX0-fCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1509492254366598534</id><published>2010-12-17T02:28:00.000-08:00</published><updated>2010-12-21T02:10:50.354-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.354-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Multiplication'/><title>How to do long multiplication step by step</title><summary>Long multiplication extends tables work so that numbers bigger than  10 can be multiplied without using a calculator. There are a number of  ways to do this.The traditional method is demonstrated in the example below. This  method is very versatile and can handle decimals as well as whole  numbers. In the box on the right you can enter your own multiplications.  Watch as the solution unfolds step</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1509492254366598534?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1509492254366598534?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-multiplication-step-by.html' title='How to do long multiplication step by step'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry><entry gd:etag='W/&quot;DEUBQX08eip7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-5512728599782382636</id><published>2010-12-16T20:16:00.000-08:00</published><updated>2010-12-21T02:10:50.372-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.372-08:00</app:edited><title>Fractions in Fractions Games</title><summary>Online Fractions Games FRACTION BALLS 1:  Identify fractions with the picture.  Each fraction ball has a fraction on it and it's your job is to guide the ball into the correct container.  MATH SPLAT:  Help save the mosquitoes in this adding fractions game.  Starts with  common denominators and the last level has unlike denominators. Multiplying Fractions Drill:  After you click the link, if you </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5512728599782382636?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5512728599782382636?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/fractions-in-fractions-games.html' title='Fractions in Fractions Games'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry><entry gd:etag='W/&quot;DEUBQX0yfSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-8457286749559123263</id><published>2010-12-16T19:55:00.000-08:00</published><updated>2010-12-21T02:10:50.395-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.395-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to do Long division tricks for kids</title><summary>"In arithmetic, long division is a standard procedure suitable for dividing simple or complex multidigit numbers. It breaks down a division problem into a series of easier steps. As in all division problems, one number, called the dividend, is divided by another, called the divisor, producing a result called the quotient. It enables computations involving arbitrarily large numbers to be performed</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/8457286749559123263?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/8457286749559123263?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-division-tricks-for-kids.html' title='How to do Long division tricks for kids'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQhyQO4SoSI/AAAAAAAAAqo/xRSaNXQktlk/s72-c/how-to-teach-long-division-step-by-step.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXo5fSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1425379611561574154</id><published>2010-12-15T01:13:00.000-08:00</published><updated>2010-12-21T02:10:50.425-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.425-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to explain long division to children?</title><summary>How to explain long division to children?Solution for 531219 ÷ 579 - with remainderStep 1Long  division works from left to right. Since 579 will not go into 5, a grey  0 has been placed over the 5 and we combine the first two digits to  make 53. In this case, 53 is still too small. A further 0 is added above  3  and a third digit is added to make 531. Note the other digits in the  original number</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1425379611561574154?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1425379611561574154?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-explain-long-division-to.html' title='How to explain long division to children?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQhyQO4SoSI/AAAAAAAAAqo/xRSaNXQktlk/s72-c/how-to-teach-long-division-step-by-step.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXo_cSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1259106849100304938</id><published>2010-12-15T01:03:00.000-08:00</published><updated>2010-12-21T02:10:50.449-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.449-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>44th International Mathematical Olympiad 2003 Problems &amp; Solutions</title><summary>A1.  S is the set {1, 2, 3, ... , 1000000}. Show that for any subset A of S with 101 elements we can find 100 distinct elements xi of S, such that the sets xi + A are all pairwise disjoint. [Note that xi + A is the set {a + xi | a is in A} ].                A2.  Find all pairs (m, n) of positive integers such that m2/(2mn2 - n3 + 1) is a positive integer.               A3.  A convex hexagon has </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1259106849100304938?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1259106849100304938?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/44th-international-mathematical.html' title='44th International Mathematical Olympiad 2003 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://1.bp.blogspot.com/_J56MO7usZSU/TQiEMXIxonI/AAAAAAAAArE/3lttEjR0nbg/s72-c/44th+International+Mathematical+Olympiad+2003+Problems+%2526+Solutions+-+A3.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXo9cSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-5479150746089794315</id><published>2010-12-15T00:49:00.000-08:00</published><updated>2010-12-21T02:10:50.469-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.469-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>43rd International Mathematical Olympiad 2002 Problems &amp; Solutions</title><summary>A1.  S is the set of all (h, k) with h, k non-negative  integers such that h + k &lt; n. Each element of S is colored red or  blue, so that if (h, k) is red and h' ≤ h, k' ≤ k, then (h', k') is also  red. A type 1 subset of S has n blue elements with different first  member and a type 2 subset of S has n blue elements with different  second member. Show that there are the same number of type 1 and </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5479150746089794315?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5479150746089794315?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/43rd-international-mathematical.html' title='43rd International Mathematical Olympiad 2002 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://2.bp.blogspot.com/_J56MO7usZSU/TQiAdpKW9II/AAAAAAAAAq0/EbvZ8cfm0o4/s72-c/43rd+International+Mathematical+Olympiad+2002+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXoyfSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-6718167730829790604</id><published>2010-12-15T00:09:00.000-08:00</published><updated>2010-12-21T02:10:50.495-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.495-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to do long division with remainders?</title><summary> Master Long Division Practice Workbook: Improve Your Math Fluency Series (Volume 8)Basic Math and Pre-Algebra Workbook For Dummies Division, Ages 7-12 (Workbook w/Music CD)  Solution for 768978 ÷ 358 - with remainderStep 1Long  division works from left to right. Since 358 will not go into 7, a grey  0 has been placed over the 7 and we combine the first two digits to  make 76. In this case, 76 is</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/6718167730829790604?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/6718167730829790604?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-division-with-remainders.html' title='How to do long division with remainders?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQhyQO4SoSI/AAAAAAAAAqo/xRSaNXQktlk/s72-c/how-to-teach-long-division-step-by-step.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXs5fyp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1678124264948410494</id><published>2010-12-14T23:48:00.000-08:00</published><updated>2010-12-21T02:10:50.527-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.527-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to teach long division with remainders?</title><summary> Master Long Division Practice Workbook: Improve Your Math Fluency Series (Volume 8)Basic Math and Pre-Algebra Workbook For Dummies Division, Ages 7-12 (Workbook w/Music CD)  Solution for 76999 ÷ 123 - with remainderStep 1Long  division works from left to right. Since 123 will not go into 7, a grey  0 has been placed over the 7 and we combine the first two digits to  make 76. In this case, 76 is </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1678124264948410494?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1678124264948410494?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-teach-long-division-with.html' title='How to teach long division with remainders?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQhyQO4SoSI/AAAAAAAAAqo/xRSaNXQktlk/s72-c/how-to-teach-long-division-step-by-step.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXs-fSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1619809284244068378</id><published>2010-12-14T23:22:00.000-08:00</published><updated>2010-12-21T02:10:50.555-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.555-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to do long division of large numbers?</title><summary>How to divide larger numbers and do long division?This is video for how to divide large numbers by doing long division with easy methods. This video describes that dividing the "Dividend" with "Divisor" to get the "Quotient" and "Remainder" if they exist. In the given example first "25" taken as the Dividend and "5" is will be the Divisor. Divide this value the multiplication is needed. For above</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1619809284244068378?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1619809284244068378?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-division-of-large.html' title='How to do long division of large numbers?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_J56MO7usZSU/TQhtWRqbfqI/AAAAAAAAAqk/6hKZ9sm5MX0/s72-c/LongDivisionAnimated.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXs8eyp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-5876791963025985956</id><published>2010-12-14T23:19:00.000-08:00</published><updated>2010-12-21T02:10:50.573-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.573-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to do long division of polynomials?</title><summary>Divide          3x3 – 5x2 + 10x – 3           by  3x + 1 This division did not come out even. What am I          supposed to do with the remainder? Think back to when you did long division with          plain numbers. Sometimes there would be a remainder; for instance, if you divide 132          by 5: ...there is a remainder of 2.          Remember how you handled that? You made a fraction, </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5876791963025985956?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5876791963025985956?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-division-of-polynomials.html' title='How to do long division of polynomials?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_J56MO7usZSU/TQhrJ7ZYkAI/AAAAAAAAAqY/TmQ6aRHnSkk/s72-c/divani03.gif' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXsycCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-8818574419812206543</id><published>2010-12-14T23:12:00.000-08:00</published><updated>2010-12-21T02:10:50.598-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.598-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='Long Division'/><title>How to do long division without a calculator?</title><summary>How to teach long division without a calculator?In  the olden days, knowing how to divide large numbers was important.  Basic long division is still good to know, so the following examples  will show you how to divide a one-digit divisor into another number, and  then how to find a remainder.Recall that the divisor in a division problem is the number  that you’re dividing by. When you’re doing </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/8818574419812206543?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/8818574419812206543?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/how-to-do-long-division-without.html' title='How to do long division without a calculator?'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://1.bp.blogspot.com/_J56MO7usZSU/TQhpN3P6rII/AAAAAAAAApg/vMXf8aef6Vk/s72-c/216886.png' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXg4eSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-5775091265250178726</id><published>2010-12-13T18:11:00.000-08:00</published><updated>2010-12-21T02:10:50.631-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.631-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>42nd International Mathematical Olympiad 2001 Problems &amp; Solutions</title><summary>A1.  ABC is acute-angled. O is its circumcenter. X is the foot of the perpendicular from A to BC. Angle C ≥ angle B + 30o. Prove that angle A + angle COX &lt; 90o.                 A2.  a, b, c are positive reals. Let a' = √(a2 + 8bc), b' = √(b2 + 8ca), c' = √(c2 + 8ab). Prove that a/a' + b/b' + c/c' ≥ 1.                 A3.  Integers are placed in each of the 441 cells of a 21 x 21  array. Each row </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5775091265250178726?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/5775091265250178726?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/42nd-international-mathematical.html' title='42nd International Mathematical Olympiad 2001 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://1.bp.blogspot.com/_J56MO7usZSU/TQbRNwwiACI/AAAAAAAAAog/6qlpThB9BkY/s72-c/42nd+International+Mathematical+Olympiad+2001+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXg-cSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-7284654711888844543</id><published>2010-12-13T18:05:00.000-08:00</published><updated>2010-12-21T02:10:50.659-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.659-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>41st International Mathematical Olympiad 2000 Problems &amp; Solutions</title><summary>A1.   AB is tangent to the circles CAMN and NMBD. M lies between C and D on  the line CD, and CD is parallel to AB. The chords NA and CM meet at P;  the chords NB and MD meet at Q. The rays CA and  DB meet at E. Prove  that PE = QE.  A2.  A, B, C are positive reals with product 1. Prove that (A - 1 + 1/B)(B - 1 + 1/C)(C - 1 + 1/A) ≤ 1.                 A3.  k is a positive real. N is an integer </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/7284654711888844543?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/7284654711888844543?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/41st-international-mathematical.html' title='41st International Mathematical Olympiad 2000 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://1.bp.blogspot.com/_J56MO7usZSU/TQbP9qrdl6I/AAAAAAAAAoQ/m7yVog5Ynm0/s72-c/41st+International+Mathematical+Olympiad+2000+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXgyfCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-3565940685405097161</id><published>2010-12-13T17:53:00.000-08:00</published><updated>2010-12-21T02:10:50.694-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.694-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>40th International Mathematical Olympiad 1999 Problems &amp; Solutions</title><summary>A1.  Find all finite sets S of at least three points in  the plane such that for all distinct points A, B in S, the perpendicular  bisector of AB is an axis of symmetry for S.                 A2.  Let n ≥ 2 be a fixed integer. Find the smallest constant C such that for all non-negative reals x1, ... , xn:      ∑i&lt;j xi xj (xi2 + xj2) ≤ C ( ∑ xi )4.  Determine when equality occurs.                 </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3565940685405097161?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3565940685405097161?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/40th-international-mathematical.html' title='40th International Mathematical Olympiad 1999 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://4.bp.blogspot.com/_J56MO7usZSU/TQbOFElJKCI/AAAAAAAAAoM/-HTScPcRZrk/s72-c/40th+International+Mathematical+Olympiad+1999+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXk6cSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-4373102086017075372</id><published>2010-12-12T18:32:00.000-08:00</published><updated>2010-12-21T02:10:50.719-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.719-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>39th International Mathematical Olympiad 1998 Problems &amp; Solutions</title><summary>A1.  In the convex quadrilateral ABCD, the diagonals AC  and BD are perpendicular and the opposite sides AB and DC are not  parallel. The point P, where the perpendicular bisectors of AB and DC  meet, is inside ABCD. Prove that ABCD is cyclic if and only if the  triangles ABP and CDP have equal areas.                 A2.  In a competition there are a contestants and b judges, where  b ≥ 3 is an </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/4373102086017075372?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/4373102086017075372?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/39th-international-mathematical.html' title='39th International Mathematical Olympiad 1998 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://2.bp.blogspot.com/_J56MO7usZSU/TQWFQqcelBI/AAAAAAAAAn4/62OIfMSrXxU/s72-c/39th+International+Mathematical+Olympiad+1998+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXk4cSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-6137711349054905567</id><published>2010-12-12T18:27:00.000-08:00</published><updated>2010-12-21T02:10:50.739-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.739-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>38th International Mathematical Olympiad 1997 Problems &amp; Solutions</title><summary>A1.  In the plane the points with integer coordinates are  the vertices of unit squares. The squares are colored alternately black  and white as on a chessboard. For any pair of positive integers m and  n, consider a right-angled triangle whose vertices have integer  coordinates and whose legs, of lengths m and n, lie along the edges of  the squares. Let S1 be the total area of the black part of </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/6137711349054905567?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/6137711349054905567?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/38th-international-mathematical.html' title='38th International Mathematical Olympiad 1997 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry><entry gd:etag='W/&quot;DEUBQXk9eCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-4026514822310231576</id><published>2010-12-11T00:28:00.000-08:00</published><updated>2010-12-21T02:10:50.760-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.760-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>37th International Mathematical Olympiad 1996 Problems &amp; Solutions</title><summary>A1.  We are given a positive integer r and a rectangular  board divided into 20 x 12 unit squares. The following moves are  permitted on the board: one can move from one square to another only if  the distance between the centers of the two squares is √r. The task is  to find a sequence of moves leading between two adjacent corners of the  board which lie on the long side.   (a)  Show that the </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/4026514822310231576?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/4026514822310231576?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/37th-international-mathematical.html' title='37th International Mathematical Olympiad 1996 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://1.bp.blogspot.com/_J56MO7usZSU/TQM1lXHAp0I/AAAAAAAAAnU/5kA53ArykGQ/s72-c/37th+International+Mathematical+Olympiad+1996+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXkzeCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-3453826161146799360</id><published>2010-12-10T03:00:00.000-08:00</published><updated>2010-12-21T02:10:50.780-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.780-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>36th International Mathematical Olympiad 1995 Problems &amp; Solutions</title><summary>A1.  Let A, B, C, D be four distinct points on a line, in  that order. The circles with diameter AC and BD intersect at X and Y.  The line XY meets BC at Z. Let P be a point on the line XY other than Z.  The line CP intersects the circle with diameter AC at C and M, and the  line BP intersects the circle with diameter BD at B and N. Prove that  the lines AM, DN, XY are concurrent.</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3453826161146799360?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3453826161146799360?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/36th-international-mathematical.html' title='36th International Mathematical Olympiad 1995 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQIHv-j5jxI/AAAAAAAAAmo/6eoPZZQWHVY/s72-c/36th+International+Mathematical+Olympiad+1995+Problems+%2526+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXY7fCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-3461322440424493467</id><published>2010-12-10T02:53:00.000-08:00</published><updated>2010-12-21T02:10:50.804-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.804-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>35th International Mathematical Olympiad 1994 Problems &amp; Solutions</title><summary>A1.  Let m and n be positive integers. Let a1, a2, ... , am be distinct elements of {1, 2, ... , n} such that whenever ai + aj ≤ n for some i, j (possibly the same) we have ai + aj = ak for some k. Prove that:      (a1 + ... + am)/m ≥ (n + 1)/2.                 A2.  ABC is an isosceles triangle with AB = AC. M is the midpoint  of BC and O is the point on the line AM such that OB is perpendicular</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3461322440424493467?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3461322440424493467?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/35th-international-mathematical.html' title='35th International Mathematical Olympiad 1994 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQIGDNEEBHI/AAAAAAAAAmk/9iO5bAKHqW4/s72-c/35th+International+Mathematical+Olympiad+1994+Problems+%26+Solutions.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXY5cSp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-9052756362788381050</id><published>2010-12-10T02:45:00.000-08:00</published><updated>2010-12-21T02:10:50.829-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.829-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>34th International Mathematical Olympiad 1993 Problems &amp; Solutions</title><summary>A1.  Let f(x) = xn + 5xn-1 + 3,  where n &gt; 1 is an integer. Prove that f(x) cannot be expressed as the  product of two non-constant polynomials with integer coefficients.                 A2.  Let D be a point inside the acute-angled triangle ABC such that ∠ADB = ∠ACB + 90o, and AC·BD = AD·BC.  (a)  Calculate the ratio AB·CD/(AC·BD).  (b)  Prove that the tangents at C to the circumcircles of ACD </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/9052756362788381050?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/9052756362788381050?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/34th-international-mathematical.html' title='34th International Mathematical Olympiad 1993 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author><media:thumbnail xmlns:media='http://search.yahoo.com/mrss/' url='http://3.bp.blogspot.com/_J56MO7usZSU/TQID6tslJgI/AAAAAAAAAmg/4Hryz2MbpaI/s72-c/34th+International+Mathematical+Olympiad+1993+Problems.jpg' height='72' width='72'/></entry><entry gd:etag='W/&quot;DEUBQXY-eip7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-1361038333490662070</id><published>2010-12-08T22:54:00.000-08:00</published><updated>2010-12-21T02:10:50.852-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.852-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>33rd International Mathematical Olympiad 1992 Problems &amp; Solutions</title><summary>A1.  Find all integers a, b, c satisfying 1 &lt; a &lt; b &lt; c such that (a - 1)(b -1)(c - 1) is a divisor of abc - 1.                 A2.  Find all functions f defined on the set of all real numbers with real values, such that f(x2 + f(y)) = y + f(x)2 for all x, y.                 A3.  Consider 9 points in space, no 4 coplanar. Each pair of  points is joined by a line segment which is colored either </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1361038333490662070?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/1361038333490662070?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/33rd-international-mathematical.html' title='33rd International Mathematical Olympiad 1992 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry><entry gd:etag='W/&quot;DEUBQXY8fCp7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-3789165498033221893</id><published>2010-12-08T22:48:00.000-08:00</published><updated>2010-12-21T02:10:50.874-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.874-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>32nd International Mathematical Olympiad 1991 Problems &amp; Solutions</title><summary>A1.  Given a triangle ABC, let I be the incenter. The  internal bisectors of angles A, B, C meet the opposite sides in A', B',  C' respectively. Prove that:      1/4 &lt; AI·BI·CI/(AA'·BB'·CC') ≤ 8/27.                 A2.  Let n &gt; 6 be an integer and let a1, a2, ... , ak be all the positive integers less than n and relatively prime to n. If      a2 - a1 = a3 - a2 = ... = ak - ak-1 &gt; 0,  prove that n</summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3789165498033221893?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3789165498033221893?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/12/32nd-international-mathematical.html' title='32nd International Mathematical Olympiad 1991 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry><entry gd:etag='W/&quot;DEUBQXc7fip7ImA9Wx9RGU8.&quot;'><id>tag:blogger.com,1999:blog-5382626319333381386.post-3268865290148069223</id><published>2010-11-24T01:39:00.000-08:00</published><updated>2010-12-21T02:10:50.906-08:00</updated><app:edited xmlns:app='http://www.w3.org/2007/app'>2010-12-21T02:10:50.906-08:00</app:edited><category scheme='http://www.blogger.com/atom/ns#' term='IMO'/><title>31st International Mathematical Olympiad 1990 Problems &amp; Solutions</title><summary>A1.  Chords AB and CD of a circle intersect at a point E  inside the circle. Let M be an interior point of the segment EB. The  tangent at E to the circle through D, E and M intersects the lines BC  and AC at F and G respectively. Find EF/EG in terms of t = AM/AB.                 A2.  Take n ≥ 3 and consider a set E of 2n-1 distinct points on a  circle. Suppose that exactly k of these points are </summary><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3268865290148069223?v=2'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5382626319333381386/posts/default/3268865290148069223?v=2'/><link rel='alternate' type='text/html' href='http://www.equationsolver.info/2010/11/31st-international-mathematical.html' title='31st International Mathematical Olympiad 1990 Problems &amp;amp; Solutions'/><author><name>Nguyen Thi Lan Phuong</name><uri>http://www.blogger.com/profile/08966362654219093206</uri><email>toantuongtac@gmail.com</email><gd:extendedProperty name='OpenSocialUserId' value='16908134740580140208'/></author></entry></feed>