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H.C.F of 3240,3600 and a third number is 36 and their L.C.M is 2^4x3^5x5^2x7^2.Find the third number.
H.C.F=36=2^2*3^2 L.C.M=2^4*3^5*5^2*7^2 3600=2^4*3^2*5^2 3240=2^2*3^4*5 Third no.= 2^2*3^2*7^2*3^3 =2^2*3^5*7^2
HCF (3240, 3600, y) = 36 LCM (2^4x3^5x5^2x7^2) = 4846100 We know that; HCF X LCM = Product of the numbers 36 X 4846100 = 3240 X 3600 X y 174459600 = 11664000 X y y = 174459600 / 11664000 y = FIND OUT ?
3600=2^4*3^2*5^23240=2^2*3^4*5H.C.F =36=2^2 *3^2Since H.C.F is the product of lowest power of common Factor , so the third Number must have (2^2 *3^2) as it`s factor .Since L.C.M is the Product of Highest power of Common Prime Factor , So the Third number must have 3^5 and 7^2 as it`s factor .Therefore , Third number =2^2*3^5*7^2
Dear student,Please find the attached solution to your problem below.3600 = 24*32*523240 = 22*34*5H.C.F = 36 = 22*32Since H.C.F is the product of lowest power of common Factor , so the third Number must have(22*32) as it`s factor.Since L.C.M is the Product of Highest power of Common Prime Factor, So the Thirdnumber must have 35 and 72 as it’s factor.Therefore, Third number =22*35*72 Hope it helps.Thanks and regards,Kushagra
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