{"id":105,"date":"2018-08-17T05:05:27","date_gmt":"2018-08-17T05:05:27","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=105"},"modified":"2019-06-09T06:44:23","modified_gmt":"2019-06-09T06:44:23","slug":"h-c-f-and-l-c-m-using-the-fundamental-theorem-of-arithmetic","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/h-c-f-and-l-c-m-using-the-fundamental-theorem-of-arithmetic\/","title":{"rendered":"How To Find The HCF And LCM Using Prime Factorisation Method"},"content":{"rendered":"
Consider two numbers 18 and 24.
\nPrime factorisation of 18 = 2 \u00d7 3 \u00d7 3
\nPrime factorisation of 24 = 2 \u00d7 2 \u00d7 2 \u00d7 3
\nSo, HCF = 2 \u00d7 3 = 6
\nLCM = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 = 72
\nProduct of HCF and LCM = 6 \u00d7 72 = 432
\nProduct of given numbers = 18 \u00d7 24 = 432
\nThe product of LCM and HCF of two natural numbers is equal to the product of the given natural numbers.
\n\u2234 Product of given numbers = HCF \u00d7 LCM of given numbers
\nFor any two positive integers:
\nTheir LCM. \u00d7 their HCF.\u00a0= Product of the number<\/p>\n
(i) LCM =\u00a0\\(\\frac{\\text{Product of the numbers}}{\\text{HCF}}\\)<\/p>\n
(ii) HCF =\u00a0\\(\\frac{\\text{Product of the numbers}}{\\text{LCM}}\\)<\/p>\n
(iii) One number =\u00a0\\(\\frac{\\text{H}\\text{.C}\\text{.F}\\text{.\u00a0 }\\!\\!\\times\\!\\!\\text{\u00a0 L}\\text{.C}\\text{.M}\\text{.}}{\\text{Other number}}\\)<\/p>\n
Find the L.C.M. and H.C.F. of the following pairs of integers by applying the Fundamental theorem of Arithmetic method i.e., using the prime factorisation method.<\/p>\n
Example 1: \u00a0 \u00a0<\/strong>26 and 91 Example 2: \u00a0 \u00a0<\/strong>1296 and 2520 Example 3: \u00a0 \u00a0<\/strong>17 and 25 Example 4: \u00a0 \u00a0<\/strong>Given that H.C.F. (306, 657) = 9,\u00a0find L.C.M. (306, 657) Example 5:<\/strong>\u00a0 \u00a0 Given that L.C.M. (150, 100) = 300, find H.C.F. (150, 100) Example 6: \u00a0 \u00a0<\/strong>The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers. Example 7:<\/strong>\u00a0 \u00a0 Explain why 7 \u00d7 11 \u00d7 13 + 13 and\u00a07 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 + 5 are composite numbers. Maths<\/a> <\/p>\n","protected":false},"excerpt":{"rendered":" Find The HCF And LCM using Prime Factorisation Method Relation between two numbers and their HCF and LCM\u00a0 Consider two numbers 18 and 24. Prime factorisation of 18 = 2 \u00d7 3 \u00d7 3 Prime factorisation of 24 = 2 \u00d7 2 \u00d7 2 \u00d7 3 So, HCF = 2 \u00d7 3 = 6 LCM … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[10,9,55912,55948,8,7],"yoast_head":"\n
\nSol.<\/strong> \u00a0 \u00a0Since, 26 = 2 \u00d7 13 and, 91 = 7 \u00d7 13
\n
\nL.C.M.<\/strong> = Product of each prime factor with highest powers. = 2 \u00d7 13 \u00d7 7 = 182.
\ni.e., L.C.M.<\/strong> (26, 91) = 182.
\nH.C.F. <\/strong>= Product of common prime factors with lowest powers. = 13.
\ni.e., H.C.F<\/strong> (26, 91) = 13.
\nProduct of given two numbers = 26 \u00d7 91 \u00a0= 2366
\nand, product of their\u00a0L.C.M.<\/strong> and H.C.F.<\/strong> = 182 \u00d7 13 = 2366
\nProduct of L.C.M and H.C.F of two given numbers = Product of the given numbers<\/p>\n
\nSol.<\/strong> \u00a0 \u00a0Since, 1296 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 = 24<\/sup> \u00d7 34<\/sup>\u00a0 <\/sup>
\nand,\u00a0\u00a0\u00a0\u00a0 2520 = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 7\u00a0= 23<\/sup> \u00d7 32<\/sup> \u00d7 5 \u00d7 7
\n
\nL.C.M.<\/strong> = Product of each prime factor with highest powers
\n= 24<\/sup> \u00d7 34<\/sup> \u00d7 5 \u00d7 7 = 45,360 <\/strong>
\ni.e., L.C.M.<\/strong> (1296, 2520) = 45,360
\nH.C.F. <\/strong>= Product of common prime factors with lowest powers\u00a0= 23<\/sup> \u00d7 32<\/sup> = 8 \u00d7 9 = 72
\ni.e., H.C.F. (1296, 2520) = 72.<\/strong>
\nProduct of given two numbers\u00a0= 1296 \u00d7 2520 = 3265920
\nand, product of their\u00a0L.C.M.<\/strong> and H.C.F.<\/strong> = 45360 \u00d7 72 = 3265920
\nL.C.M. (1296, 2520) \u00d7 H.C.F. (1296, 2520)
\n= 1296 \u00d7 2520 = 3265920<\/p>\n
\nSol.<\/strong> \u00a0 \u00a0Since, \u00a0 17 = 17
\nand,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 25 = 5 \u00d7 5 = 52<\/sup>
\nL.C.M.<\/strong> = 17 \u00d7 52<\/sup> = 17 \u00d7 25 = 425
\nand, H.C.F.<\/strong> = Product of common prime factors\u00a0 with lowest powers\u00a0= 1, as given numbers do not have any common prime factor.
\nThe given numbers 17 and 25 do not have any common prime factor. Such numbers are called co-prime numbers and their H.C.F. is always equal to 1 (one), whereas their L.C.M. is equal to the product of the numbers.
\nBut in case of two co-prime numbers also, the product of the numbers is always equal to the product of their L.C.M. and their H.C.F.
\nAs, in case of co-prime numbers 17 and 25;
\nH.C.F. = 1; L.C.M. = 17 \u00d7 25 = 425;
\nproduct of numbers = 17 \u00d7 25 = 425
\nand product of their H.C.F. and L.C.M.\u00a0= 1 \u00d7 425 = 425.<\/p>\n
\nSol.<\/strong>\u00a0 \u00a0 \u00a0H.C.F. (306, 657) = 9 means H.C.F. of
\n306 and 657 = 9
\nRequired L.C.M. (306, 657) means required L.C.M. of 306 and 657.
\nFor any two positive integers;
\ntheir L.C.M. =\u00a0\\(\\frac{\\text{Product of the numbers}}{\\text{H}\\text{.C}\\text{.F}\\text{.}}\\)
\ni.e., L.C.M. (306, 657) = \\(\\frac{306\\times 657}{9}\\)\u00a0 = 22,338.<\/p>\n
\nSol.<\/strong>\u00a0 \u00a0 \u00a0L.C.M. (150, 100) = 300
\n\u21d2 L.C.M. of 150 and 100 = 300
\nSince, the product of number 150 and 100
\n= 150 \u00d7 100
\nAnd, we know :
\nH.C.F. (150, 100) =\u00a0\\(\\frac{\\text{Product of 150 and 100}}{L.C.M\\text{.(150,100)}}\\)
\n=\u00a0\\(\\frac{150\\times 100}{300}\\)\u00a0= 50.<\/p>\n
\nSol.<\/strong>\u00a0 \u00a0 \u00a0Since, the product of two numbers
\n= Their H.C.F. \u00d7 Their L.C.M.
\n\u21d2 One no. \u00d7 other no. = H.C.F. \u00d7 L.C.M.
\n\u21d2 Other no. = \\(\\frac{12\\times 240}{48}\\) = 60.<\/p>\n
\nSol.<\/strong>\u00a0 \u00a0 Since,
\n7 \u00d7 11 \u00d7 13 + 13 = 13 \u00d7 (7 \u00d7 11 + 1)
\n= 13 \u00d7 78 = 13 \u00d7 13 \u00d7 3 \u00d7 2;
\nthat is, the given number has more than two factors and it is a composite number.
\nSimilarly, 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 + 5
\n= 5 \u00d7 (7 \u00d7 6 \u00d7 4 \u00d7 3 + 1)
\n= 5 \u00d7 505 = 5 \u00d7 5 \u00d7 101
\n\u2234\u00a0The given no. is a composite number.<\/p>\n