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lcm and hcf by factorisation


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Find the hcf and lcm of 12,15,21
Use prime factorisation
12 = [2^2]*[3]
15=[3]*[5]
21=[3]*[7]

For HCF, you have to find the  common factors
and 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

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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

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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]

LCM(48,72,108)=[2^4]*[3^3]=16*27=432

The lights will change simultaneously

432 seconds after 8:20:00 hours
using long division
432 seconds =7*60 +12 =7minutes 12seconds
after
 
8:20:00 hours
or at 8:27:12 hours

other questions and problems:


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