{"id":10387,"date":"2022-04-05T17:30:04","date_gmt":"2022-04-05T12:00:04","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10387"},"modified":"2022-06-04T19:39:41","modified_gmt":"2022-06-04T14:09:41","slug":"find-the-h-c-f-and-l-c-m-of-the-following-integers-by-applying-prime-factorisation-method","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-h-c-f-and-l-c-m-of-the-following-integers-by-applying-prime-factorisation-method\/","title":{"rendered":"Find the H.C.F and L.C.M of the following integers by applying prime factorisation method."},"content":{"rendered":"
Question<\/strong> : Find the H.C.F and L.C.M of the following integers by applying prime factorisation method.<\/p>\n (i)\u00a0 12, 15, 21<\/p>\n (ii)\u00a0 17, 23, 29<\/p>\n (iii)\u00a0 8, 9 and 25<\/p>\n Solution<\/strong> :<\/p>\n (i)\u00a0<\/strong>12, 15, 21<\/p>\n 12 = \\(2 \\times 2 \\times 3\\)<\/p>\n 15 = \\(3 \\times 5\\)<\/p>\n 21 = \\(3 \\times 7\\)<\/p>\n Here 3 is a common prime factor of the given numbers.<\/p>\n Hence, HCF = 3<\/p>\n H.C.F. (12, 15, 21) = 3<\/strong><\/p>\n L.C.M is the product of the prime factors\u00a0 \\(2 \\times 2 \\times 3 \\times 7 \\times 5\\)<\/p>\n The common factor 3 is repeated three times, so we write 3 only one times in multiplication..<\/p>\n L.C.M. (12, 15, 21) = 420<\/strong><\/p>\n (ii)\u00a0<\/strong>17, 23, 29<\/p>\n H.C.F. There is no common factor as 17, 23, 29 they are prime. Hence, H.C.F is 1.<\/p>\n L.C.M. is the product of all prime factors 17, 23 and 29.<\/p>\n \\(17 \\times 23 \\times 29\\) = 11339<\/p>\n Hence, H.C.F. (17, 23, 29) = 1.<\/strong><\/p>\n L.C.M (17, 23, 29) = 11339<\/strong><\/p>\n (iii)\u00a0<\/strong>8, 9 and 25<\/p>\n First we write the prime factorisation of each of the given numbers.<\/p>\n 8 = \\(2 \\times 2 \\times\\) = \\(2^3\\),<\/p>\n 9 = \\(3 \\times 3\\) = \\(3^2\\)<\/p>\n 25 = \\(5 \\times 5\\) = \\(5^2\\)<\/p>\n \\(\\therefore\\)\u00a0 L.C.M = \\(2^2 \\times 3^2 \\times 5^2\\) = \\(8 \\times 9 \\times 25\\) = 1800<\/strong><\/p>\n and H.C.F = 1.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Question : Find the H.C.F and L.C.M of the following integers by applying prime factorisation method. (i)\u00a0 12, 15, 21 (ii)\u00a0 17, 23, 29 (iii)\u00a0 8, 9 and 25 Solution : (i)\u00a012, 15, 21 12 = \\(2 \\times 2 \\times 3\\) 15 = \\(3 \\times 5\\) 21 = \\(3 \\times 7\\) Here 3 is a …<\/p>\n