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Re: Reverse of LCM / HCF [#permalink]
thanks for the input
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Re: Reverse of LCM / HCF [#permalink]
VeritasPrepKarishma wrote:
Also remember, if a and b are 2 numbers whose HCF and LCM is given,
a*b = HCF*LCM
(Try to figure out why this must be true.)


karishma
I never appreciated the formulas for LCM and HCF of the fractions. I just don't seem to get these formulas, can you throw some light, how?

LCM of a fraction is - LCM of numerator/HCF of denominator. 21/2 is the LCM of 3/4 and 7/6

HCF of a fraction is - HCF of numerator/LCM of denominator. 1/12 is the HCF of 3/4 and 7/6

This looked daunting even during my school days until I started using LCM to compare fractions wherein it did make sense. But not the LCM of the fractions. :)
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Re: Reverse of LCM / HCF [#permalink]
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gmat1220 wrote:
VeritasPrepKarishma wrote:
Also remember, if a and b are 2 numbers whose HCF and LCM is given,
a*b = HCF*LCM
(Try to figure out why this must be true.)


karishma
I never appreciated the formulas for LCM and HCF of the fractions. I just don't seem to get these formulas, can you throw some light, how?

LCM of a fraction is - LCM of numerator/HCF of denominator. 21/2 is the LCM of 3/4 and 7/6

HCF of a fraction is - HCF of numerator/LCM of denominator. 1/12 is the HCF of 3/4 and 7/6

This looked daunting even during my school days until I started using LCM to compare fractions wherein it did make sense. But not the LCM of the fractions. :)


LCM/HCF is an important concept as basis for other concepts but LCM/HCF of fractions isn't that important. Over the years, I remember using this concept very rarely. Nevertheless, of course no harm in having a clear understanding.

Algebraic approach

Consider 2 fractions a/b and c/d in their lowest form, their LCM (L1/L2) and HCF (H1/H2) (also in their lowest forms)
LCM should be divisible by both numbers so
L1/L2 is divisible by a/b. This implies L1*b/L2*a is an integer. Since a/b and L1/L2 are in their lowest form, L1 must be divisible by a and b must be divisible by L2.
Similarly, L1 must be divisible by c and d must be divisible by L2.
L1, the numerator of LCM, must be divisible by both a and c and hence should be the LCM of a and c, the numerators. (L1 cannot be just any multiple of a and c; it must be the lowest common multiple so that L1/L2 is the Lowest Multiple of the two fractions)
b and d both must be divisible by L2 and hence L2 must be their HCF. (Not just any common factor but the highest common factor so that L1/L2 is the lowest multiple possible)

Using similar reasoning, you can figure out why we find HCF of fractions the way we do.

Now let me give you some feelers. They are more important than the algebraic explanation above. They build intuition.

Let me remind you first that LCM is the lowest common multiple. It is that smallest number which is divisible by both the given numbers.
Say, I have two fractions: 1/4 and 1/2. What is their LCM? It's 1/2 because 1/2 is the smallest fraction which is divisible by both 1/2 and 1/4. (If this is tricky to see, think about their equivalents in decimal form 1/2 = 0.50 and 1/4 = 0.25. You can see that 0.50 is the smallest common multiple they have)

But 1/2 = 2/4. LCM of 2/4 and 1/4 will obviously be 2/4....

What is HCF? It is that greatest number which is common between the two fractions. Again, let's take 1/2 and 1/4. What is greatest common fraction between them? 1/4 (Note that 1/2 and 1/4 are both divisible by other fractions too e.g. 1/8, 1/24 etc but 1/4 is the greatest such common fraction)

On the same lines, what will be the LCM of 2/3 and 1/8. We know that 2/3 = 16/24 and 1/8 = 3/24. what do you think their LCM will be?
16*3/24 = 48/24 = 2

Also, think what will be the HCF of 2/3 and 1/8. We know that 2/3 = 16/24 and 1/8 = 3/24. What is common between the two fractions? 1/24

LCM is a fraction greater than (or equal to) both the fractions. When you take the LCM of the numerator and HCF of the denominator, you are making a fraction greater than (or equal to) either one of the numbers.

HCF is a fraction smaller than (or equal to) both the fractions. When you take the HCF of the numerator and LCM of the denominator, you are making a fraction smaller than (or equal to) either one of the numbers.
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Re: Reverse of LCM / HCF [#permalink]
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True. Thanks!!! I also get the following two inferences. Pls verify:

1. Lcm is the multiple of hcf.
2. Never trust lcm of two numbers alone for deriving the number properties of two numbers. I mean lcm will include primes from both numbers. Hence the prime can belong to either one. You cant say which one.

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Re: Reverse of LCM / HCF [#permalink]
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gmat1220 wrote:
Karishma
True. Thanks!!! I also get the following two inferences. Pls verify:

1. Lcm is the multiple of hcf.
2. Never trust lcm of two numbers alone for deriving the number properties of two numbers. I mean lcm will include primes from both numbers. Hence the prime can belong to either one. You cant say which one.

Posted from my mobile device


Yes, you are right.
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Re: Reverse of LCM / HCF [#permalink]
VeritasPrepKarishma wrote:
rxs0005 wrote:
hi all

How can we find the different numbers associated with an GCD , HCF

for eg :

if we have the GCD is 30 of 2 numbers how can we back trace the possible numbers like

6 , 5 : 10 , 3 etc


and if we have HCF as 5 is there a way we can find potential number that have HCF 5


GCD/HCF (As pointed out above, they are the same) is the highest common factor so if HCF = 30, both numbers will definitely have 30 as a factor. The two numbers will be 30a and 30b where a and b will have no common factors (because if they did have a common factor, then HCF would have been 30*the common factor)
So numbers could be (30*4 and 30*5) but not (30*4 and 30*8).

LCM is the lowest common multiple i.e. it contains both the numbers. If LCM = 30 = 2*3*5, the numbers could be
1 and 30 (Start with 1 and the LCM)
2 and 15
2 and 30
6 and 5
etc

The point to remember is that both numbers must be made of only a single 2 and/or a single 3 and/or a single 5.
So the numbers could be (2*3 and 3*5) but not (2*2*3 and 3*7)

Also remember, if a and b are 2 numbers whose HCF and LCM is given,
a*b = HCF*LCM
(Try to figure out why this must be true.)


hi Karishma, I don't get why So numbers could be (30*4 and 30*5) but not (30*4 and 30*8)...aren't 4 nor 8 the prime factor so i supposed they should not shown up in the HCF ?
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Re: Reverse of LCM / HCF [#permalink]
miweekend wrote:
hi Karishma, I don't get why So numbers could be (30*4 and 30*5) but not (30*4 and 30*8)...aren't 4 nor 8 the prime factor so i supposed they should not shown up in the HCF ?

If numbers are 30*4 and 30*8, then HCF should be 30*4 which is HCF = 120 and both numbers will definitely have 120 as a factor. The corresponding numbers will be 120a and 120b where a and b will have no common factors.

The same doesn't exist in 30*4 and 30*5 because both numbers definitely have 30 as a factor.
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Re: Reverse of LCM / HCF [#permalink]
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miweekend wrote:

hi Karishma, I don't get why So numbers could be (30*4 and 30*5) but not (30*4 and 30*8)...aren't 4 nor 8 the prime factor so i supposed they should not shown up in the HCF ?


I think Varun explained why numbers cannot be 30*4 and 30*8 quite well. Let me add a little bit of theory here:

What is GCD? It is the greatest common divisor (or highest common factor). This means that if GCD of two numbers, a and b, is 30, the greatest common divisor of a and b will be 30 i.e. once I divide a and b by 30, they will not have any other common factor left (except 1). Say, a = 120 and b = 150. Their GCD is 30.
a/30 = 4
b/30 = 5
4 and 5 have no common factor except 1.

If a = 120 and b = 240,
a/30 = 4
b/30 = 8
They do have a common factor 4. Hence 30 is not their GCD. Their GCD is 120.
a/120 = 1
b/120 = 2
Now we see that a and b have no common factor.
So basically, GCD is the greatest number that divides both numbers. Once you divide by GCD, you are left with co-primes i.e. numbers which have no common factor except 1.
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Re: Reverse of LCM / HCF [#permalink]
Karishma.. Thanks for such a clear explanation.. :)
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