lcm and hcf by factorisation

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--------------------------------------------------------------------- Find the hcf and lcm of 12,15,21Use
factorisationprime12 = [2^2]*[3] 15=[3]*[5] 21=[3]*[7] For HCF, you have to find the factorscommonand then take the product of the lowest power of those common factors across the numbers Here only [3^1] is the common factor hence HCF(12,15,21)= 3^1 = 3 For LCM you have to consider all the factors and take the product of the highest power of all the factors LCM(12,15,21) =[2^2]*[3^1]*[5^1]*[7^1]= 420 -------------------------------------- Find the HCF of 70,105,175 70=2*5*7 105=5*3*7 175=[5^2]*7 here 5 and 7 are the common factors with lowest power [5^1] and[7^1] HCF(70,105,175)=5*7 = 35 ----------------------------- The traffic lights at three different road crossing change after every 48 sec, 72sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hours, when will they change simultaneously again We have to find LCM(48,72,108) 48=[2^4]*[3] 72=[2^3]*[3^2] 108=[2^2]*[3^3] 8:20:00 hoursLCM(48,72,108)=[2^4]*[3^3]=16*27=432The lights will change simultaneously 432 seconds after 8:20:00 hours using long division 432 seconds =7*60 +12 =7minutes 12seconds after or at 8:27:12 hours other questions and problems: * integral
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