{"id":86,"date":"2023-05-04T10:00:45","date_gmt":"2023-05-04T04:30:45","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=86"},"modified":"2023-05-05T09:16:24","modified_gmt":"2023-05-05T03:46:24","slug":"euclid-division-algorithm","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/euclid-division-algorithm\/","title":{"rendered":"What is Euclid Division Algorithm"},"content":{"rendered":"<h2><strong>What is Euclid Division Algorithm<\/strong><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-12412\" src=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/euclid-division-algorithm.jpg\" alt=\"\" width=\"638\" height=\"359\" srcset=\"https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/euclid-division-algorithm.jpg 638w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/euclid-division-algorithm-300x169.jpg 300w, https:\/\/www.aplustopper.com\/wp-content\/uploads\/2016\/08\/euclid-division-algorithm-358x200.jpg 358w\" sizes=\"(max-width: 638px) 100vw, 638px\" \/><\/p>\n<p><strong>Euclid\u2019s Division Lemma:<\/strong><br \/>\nFor any two positive integers <strong>a <\/strong>and <strong>b<\/strong>, there exist unique integers q and r satisfying a = bq + r, where 0\u00a0\u2264\u00a0r &lt; b.<br \/>\n<strong>For Example<\/strong><br \/>\n(i) Consider number 23 and 5, then:<br \/>\n23 = 5 \u00d7 4 + 3<br \/>\nComparing with a = bq + r; we get:<br \/>\na = 23, b = 5, q = 4, r = 3<br \/>\nand 0 \u2264\u00a0r &lt; b (as 0 \u2264\u00a03 &lt; 5).<br \/>\n(ii)\u00a0 Consider positive integers 18 and 4.<br \/>\n18 = 4 \u00d7 4 + 2<br \/>\n\u21d2 For 18 (= a) and 4(= b) we have q = 4,<br \/>\nr = 2 and \u00a00 \u2264\u00a0r &lt; b.<br \/>\nIn the relation a = bq + r, where 0 \u2264\u00a0r &lt; b is nothing but a statement of the long division of number a by number b in which q is the quotient obtained and r is the remainder.<br \/>\nThus, dividend = divisor \u00d7 quotient + remainder \u21d2\u00a0a = bq + r<br \/>\n<strong>H.C.F. (Highest Common Factor) <\/strong><br \/>\nThe H.C.F. of two or more positive integers is the largest positive integer that divides each given positive number completely.<br \/>\ni.e., if positive integer <strong>d<\/strong> divides two positive integers <strong>a<\/strong> and <strong>b<\/strong> then the H.C.F. of <strong>a<\/strong> and <strong>b<\/strong> is <strong>d<\/strong>.<br \/>\n<strong>For Example<\/strong><br \/>\n(i)\u00a0\u00a0 14 is the largest positive integer that divides 28 and 70 completely; therefore H.C.F. of 28 and 70 is 14.<br \/>\n(ii) H.C.F. of 75, 125 and 200 is 25 as 25 divides each of 75, 125 and 200 completely and so on.<br \/>\n<strong>Using Euclid\u2019s Division Lemma For Finding H.C.F.<\/strong><br \/>\nConsider positive integers 418 and 33.<br \/>\n<strong>Step-1<\/strong><br \/>\nTaking bigger number (418) as <strong>a<\/strong> and smaller number (33) as <strong>b<\/strong><br \/>\nexpress the numbers as a = bq + r<br \/>\n\u21d2 418 = 33 \u00d7 12 + 22<strong>\u00a0<\/strong><br \/>\n<strong>Step-2<\/strong><br \/>\nNow taking the divisor 33 and remainder 22; apply the Euclid\u2019s division algorithm to get:<br \/>\n33 = 22 \u00d7 1 + 11 \u00a0\u00a0[Expressing as a = bq + r]<br \/>\n<strong>Step-3<\/strong><br \/>\nAgain with new divisor 22 and new remainder 11; apply the Euclid\u2019s division algorithm to get:<br \/>\n22 = 11 \u00d7 2 + 0<br \/>\n<strong>Step-4<\/strong><br \/>\nSince, the remainder = 0 so we cannot proceed further.<br \/>\n<strong>Step-5<\/strong><br \/>\nThe last divisor is 11 and we say H.C.F. of 418 and 33 = 11<br \/>\n<strong>Verification :<\/strong><br \/>\n<strong>(i) Using factor method:<\/strong><br \/>\n\u2234\u00a0Factors of 418 = 1, 2, 11, 19, 22, 38, 209 and 418 and,<br \/>\nFactor of 33 = 1, 3, 11 and 33.<br \/>\nCommon factors = 1 and 11<br \/>\n\u21d2 Highest common factor = 11 i.e., H.C.F. = 11<br \/>\n<strong>(ii)\u00a0 Using prime factor method:<\/strong><br \/>\nPrime factors of 418 = 2, 11 and 19.<br \/>\nPrime factors of 33 = 3 and 11.<br \/>\n\u2234\u00a0<strong>H.C.F.<\/strong> = Product of all common prime factors\u00a0 = 11.<br \/>\nFor any two positive integers a and b which can be expressed as a = bq + r, where 0 \u2264 r &lt; b, the, H.C.F. of (a, b) = H.C.F. of (q, r) and so on. For number 418 and 33<br \/>\n418 = 33 \u00d7 12 + 22<br \/>\n33 = 22 \u00d7 1 + 11<br \/>\nand 22 = 11 \u00d7 2 + 0<br \/>\n\u21d2 H.C.F. of (418, 33) = H.C.F. of (33, 22)<br \/>\n\u21d2 H.C.F. of (22, 11) = 11.<\/p>\n<h2 style=\"text-align: left;\"><strong>Euclid Division Algorithm Example Problems With Solutions<\/strong><\/h2>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/G48nSOOSDp4?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 1: \u00a0 \u00a0<\/strong>Using Euclid\u2019s division algorithm, find the H.C.F. of 135 and 225<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0Starting with the larger number i.e., 225, we get:<br \/>\n225 = 135 \u00d7 1 + 90<br \/>\nNow taking divisor 135 and remainder 90, we get<br \/>\n135 = 90 \u00d7 1 + 45<br \/>\nFurther taking divisor 90 and remainder 45, we get<br \/>\n90 = 45 \u00d7 2 + 0<br \/>\n\u2234\u00a0<strong>Required H.C.F. = 45 \u00a0<\/strong><\/p>\n<p><strong>Example 2: \u00a0\u00a0<\/strong>Using Euclid\u2019s division algorithm, find the H.C.F. of 196 and 38220<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0Starting with larger number 38220, we get:<br \/>\n38220 = 196 \u00d7 195 + 0<br \/>\nSince, the remainder is 0<br \/>\n\u2234\u00a0<strong>H.C.F. = 196\u00a0\u00a0\u00a0 <\/strong><\/p>\n<p><strong>Example 3:<\/strong>\u00a0 \u00a0 Using Euclid\u2019s division algorithm, find the H.C.F. of (iii) 867 and 255<br \/>\n<strong>Sol.<\/strong> \u00a0 \u00a0Given number are 867 and 255<br \/>\n867 = 255 \u00d7 3 + 102 \u00a0<strong>(Step-1)<\/strong><br \/>\n255 = 102 \u00d7 2 + 51 \u00a0<strong>(Step-2)<\/strong><br \/>\n102 = 51 \u00d7 2 + 0 \u00a0<strong>(Step-3)<\/strong><br \/>\n\u2234\u00a0<strong>H.C.F. = 51<\/strong><\/p>\n<p><iframe loading=\"lazy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/huXU0lOM9SQ?feature=oembed\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/p>\n<p><strong>Example 4: \u00a0 \u00a0<\/strong>Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some integer.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>According to Euclid\u2019s division lemma, if a and b are two positive integers such that a is greater than b; then these two integers can be expressed as<br \/>\na = bq + r; where 0 \u2264 r &lt; b<br \/>\nNow consider<br \/>\nb = 2; then a = bq + r will reduce to<br \/>\na = 2q + r; where 0 \u2264 \u00a0r &lt; 2,<br \/>\ni.e., r = 0 or r = 1<br \/>\nIf\u00a0\u00a0\u00a0 r = 0, a = 2q + r \u21d2 a = 2q<br \/>\ni.e., a is even<br \/>\nand, if \u00a0 \u00a0r = 1, a = 2q + r \u21d2 a = 2q + 1<br \/>\ni.e., a is add;<br \/>\nas if the integer is not even; it will be odd.<br \/>\nSince, a is taken to be any positive integer so it is applicable to the every positive integer that when it can be expressed as<br \/>\na = 2q<br \/>\n\u2234\u00a0<strong>a<\/strong> is even and when it can expressed as<br \/>\na = 2q + 1; a is odd.<br \/>\n<strong>Hence the required result.<\/strong><\/p>\n<p><strong>Example 5: \u00a0 \u00a0<\/strong>Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Let <strong>a<\/strong> is <strong>b<\/strong> be two positive integers in which <strong>a<\/strong> is greater than <strong>b<\/strong>. According to Euclid\u2019s division algorithm; <strong>a<\/strong> and <strong>b<\/strong> can be expressed as<br \/>\na = bq + r, where q is quotient and r is remainder and 0 \u2264 r &lt; b.<br \/>\nTaking b = 4, we get: a = 4q + r,<br \/>\nwhere 0 \u2264 r &lt; 4 i.e., r = 0, 1, 2 or 3<br \/>\n<strong>r = 0<\/strong>\u00a0\u21d2 a = 4q, which is divisible by 2 and so is <strong>even<\/strong>.<br \/>\n<strong>r = 1<\/strong>\u00a0\u21d2 a = 4q + 1, which is not divisible by 2 and so is <strong>odd<\/strong>.<br \/>\n<strong>r = 2<\/strong>\u00a0\u21d2 q = 4q + 2, which is divisible by 2 and so is <strong>even<\/strong>.<br \/>\nand <strong>r = 3<\/strong> \u00de q = 4q + 3, which is not divisible by 2 and so is <strong>odd<\/strong>.<br \/>\n<strong>Any positive odd integer is of the form<\/strong><br \/>\n<strong>4q + 1 or 4q + 3; <\/strong>where q is an integer.<br \/>\n<strong>Hence the required result.<\/strong><\/p>\n<p><strong>Example 6: \u00a0\u00a0<\/strong>Show that one and only one out of n; n + 2 or\u00a0n + 4 is divisible by 3, where <strong>n<\/strong> is any positive integer.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Consider any two positive integers <strong>a<\/strong> and <strong>b<\/strong> such that <strong>a<\/strong> is greater than <strong>b<\/strong>, then according to Euclid\u2019s division algorithm:<br \/>\na = bq + r; where q and r are positive integers and 0 \u2264\u00a0\u00a0r &lt; b<br \/>\nLet a = n and b = 3, then<br \/>\na = bq + r \u21d2\u00a0n = 3q + r; where 0 \u2264\u00a0\u00a0r &lt; 3.<br \/>\n<strong>r = 0<\/strong>\u00a0\u21d2 n = 3q + 0 = 3q<br \/>\n<strong>r = 1<\/strong>\u00a0\u21d2 n = 3q + 1\u00a0 and <strong>r = 2<\/strong>\u00a0\u21d2 n = 3q + 2<br \/>\nIf n = 3q; <strong>n is divisible by 3<\/strong><br \/>\nIf n = 3q + 1; then n + 2 = 3q + 1 + 2<br \/>\n= 3q + 3; which is divisible by 3<br \/>\n\u21d2\u00a0<strong>n + 2 is divisible by 3<\/strong><br \/>\nIf n = 3q + 2; then n + 4 = 3q + 2 + 4<br \/>\n= 3q + 6; which is divisible by 3<br \/>\n\u21d2\u00a0<strong>n + 4 is divisible by 3<\/strong><br \/>\nHence, if <strong>n<\/strong> is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3. <strong>\u00a0\u00a0 <\/strong><br \/>\n<strong>Hence the required result.<\/strong><\/p>\n<p><strong>Example 7:<\/strong>\u00a0 \u00a0 Show that any positive integer which is of the form 6q + 1 or 6q + 3 or 6q + 5 is odd, where q is some integer.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 If <strong>a<\/strong> and <strong>b<\/strong> are two positive integers such that <strong>a<\/strong> is greater than <strong>b<\/strong>; then according to Euclid\u2019s division algorithm; we have<br \/>\na = bq + r; where q and r are positive integers and 0 \u2264\u00a0r &lt; b.<br \/>\nLet b = 6, then<br \/>\na = bq + r \u21d2\u00a0<strong>a = 6q + r<\/strong>; where 0 \u2264\u00a0r &lt; 6.<br \/>\nWhen r = 0 \u21d2\u00a0a = 6q + 0 = 6q;<br \/>\n<strong>which is even integer<\/strong><br \/>\nWhen\u00a0\u00a0\u00a0 r = 1 \u21d2\u00a0a = 6q + 1<br \/>\n<strong>which is odd integer <\/strong><br \/>\nWhen\u00a0\u00a0\u00a0 r = 2 \u21d2\u00a0a = 6q + 2\u00a0 <strong>which is even.<\/strong><br \/>\nWhen\u00a0\u00a0\u00a0 r = 3 \u21d2\u00a0a = 6q + 3\u00a0 <strong>which is odd.<\/strong><br \/>\nWhen\u00a0\u00a0\u00a0 r = 4 \u21d2\u00a0a = 6q + 4\u00a0 <strong>which is even.<\/strong><br \/>\nWhen\u00a0\u00a0\u00a0 r = 5 \u21d2\u00a0a = 6q + 5\u00a0 <strong>which is odd.<\/strong><br \/>\nThis verifies that when r = 1 or 3 or 5; the integer obtained is 6q + 1 or 6q + 3 or 6q + 5 and each of these integers is a positive odd number.<br \/>\n<strong>Hence the required result.<\/strong><\/p>\n<p><strong>Example 8: \u00a0 \u00a0<\/strong>Use Euclid\u2019s Division Algorithm to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<br \/>\n<strong>Sol. \u00a0 \u00a0<\/strong>Let <strong>a <\/strong>and <strong>b<\/strong> are two positive integers such that <strong>a<\/strong> is greater than <strong>b<\/strong>; then:<br \/>\na = bq + r; where q and r are also positive integers and 0 \u2264\u00a0r &lt; b<br \/>\nTaking b = 3, we get:<br \/>\na = 3q + r; where 0 \u2264\u00a0r &lt; 3<br \/>\n\u21d2 The value of positive integer <strong>a<\/strong> will be<br \/>\n3q + 0, 3q + 1 or 3q + 2<br \/>\ni.e., 3q, 3q + 1 or 3q + 2.<br \/>\nNow we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m, or 3m + 1 for some integer m.<br \/>\n<strong>Square of 3q <\/strong>= (3q)<sup>2 <\/sup><br \/>\n= 9q<sup>2<\/sup> = 3(3q<sup>2<\/sup>) = 3m; 3 where m is some integer.<br \/>\n<strong>Square of 3q + 1<\/strong> = (3q + 1)<sup>2<\/sup><br \/>\n= 9q<sup>2<\/sup> + 6q + 1<br \/>\n= 3(3q<sup>2<\/sup> + 2q) + 1 = 3m + 1 for some integer m.<br \/>\n<strong>Square of 3q + 2<\/strong> = (3q + 2)<sup>2<\/sup><br \/>\n= 9q<sup>2<\/sup> + 12q + 4<br \/>\n= 9q<sup>2<\/sup> + 12q + 3 + 1<br \/>\n= 3(3q<sup>2<\/sup> + 4q + 1) + 1 = <strong>3m + 1<\/strong> for some integer m.<br \/>\nThe square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<br \/>\n<strong>Hence the required result.<\/strong><\/p>\n<p><strong>Example 9:<\/strong>\u00a0 \u00a0 Use Euclid\u2019s Division Algorithm to show that the cube of any positive integer is either of the 9m, 9m + 1 or 9m + 8 for some integer\u00a0 m.<br \/>\n<strong>Sol.<\/strong>\u00a0 \u00a0 Let <strong>a<\/strong> and <strong>b<\/strong> be two positive integers such that <strong>a<\/strong> is greater than <strong>b<\/strong>; then:<br \/>\n<strong>a = b<\/strong>q + r; where q and r are positive integers and 0 \u2264\u00a0r &lt; b.<br \/>\nTaking b = 3, we get:<br \/>\na = 3q + r; where 0 \u2264\u00a0r &lt; 3<br \/>\n\u21d2 Different values of integer <strong>a<\/strong> are<br \/>\n3q, 3q + 1 or 3q + 2.<br \/>\n<strong>Cube of 3q<\/strong> = (3q)<sup>3<\/sup> = 27q<sup>3<\/sup> = 9(3q<sup>3<\/sup>) = 9m; where <strong>m<\/strong> is some integer.<br \/>\n<strong>Cube of 3q + 1<\/strong> = (3q + 1)<sup>3<\/sup><br \/>\n= (3q)<sup>3<\/sup> + 3(3q)<sup>2<\/sup> \u00d71 + 3(3q) \u00d7 1<sup>2<\/sup> + 1<sup>3<\/sup><br \/>\n[Q (q + b)<sup>3<\/sup> = a<sup>3<\/sup> + 3a<sup>2<\/sup>b + 3ab<sup>2<\/sup> + 1]<br \/>\n= 27q<sup>3<\/sup> + 27q<sup>2<\/sup> + 9q + 1<br \/>\n= 9(3q<sup>3<\/sup> + 3q<sup>2<\/sup> + q) + 1<br \/>\n= <strong>9m + 1<\/strong>; where m is some integer.<br \/>\n<strong>Cube of 3q + 2<\/strong> = (3q + 2)<sup>3<\/sup><br \/>\n= (3q)<sup>3<\/sup> + 3(3q)<sup>2<\/sup> \u00d7 2 + 3 \u00d7 3q \u00d7 2<sup>2<\/sup> + 2<sup>3<\/sup><br \/>\n= 27q<sup>3<\/sup> + 54q<sup>2<\/sup> + 36q + 8<br \/>\n= 9(3q<sup>3<\/sup> + 6q<sup>2<\/sup> + 4q) + 8<br \/>\n= <strong>9m + 8<\/strong>; where m is some integer.<br \/>\nCube of any positive integer is of the form 9m or 9m + 1 or 9m + 8.<br \/>\n<strong>Hence the required result.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is Euclid Division Algorithm Euclid\u2019s Division Lemma: For any two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where 0\u00a0\u2264\u00a0r &lt; b. For Example (i) Consider number 23 and 5, then: 23 = 5 \u00d7 4 + 3 Comparing with a = bq + [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[11,13,12,7],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v17.8 (Yoast SEO v17.8) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>What is Euclid Division Algorithm - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/euclid-division-algorithm\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"What is Euclid Division Algorithm\" \/>\n<meta property=\"og:description\" content=\"What is Euclid Division Algorithm Euclid\u2019s Division Lemma: For any two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where 0\u00a0\u2264\u00a0r &lt; b. 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