LCM AND HCF

[Saumyojit]

if lcm of two nos is 180 and hcf is 15 . THen find the possible sets of two nos.
My question is why the first basic pair of two nos shall be 15 and 180 itself bcoz lcm of 15 and 180 is 180 and hcf is 15 itself it just common sense or there are other intutitive explanation avaialble .
@Dr.Peterson

 

[Dr.Peterson]

This will be true any time you are given two numbers, one of which is a multiple of the other: the smaller one is their HCF, and the larger one is their LCM. And since the HCF of any two numbers is a factor of their LCM, any problem like this will have a trivial solution consisting of the two numbers themselves.

 

[Saumyojit]

but here i have been given lcm and hcf not two original nos but it happens to be lcm and hcf are multiples of each other so thats why trivially 15 and 180 belongs to set of nos whose lcm and hcf is 15 and 180

 

[pka]

but here i have been given lcm and hcf not two original nos but it happens to be lcm and hcf are multiples of each other so thats why trivially 15 and 180 belongs to set of nos whose lcm and hcf is 15 and 180

Factored: \(15=3\cdot 5~\&~180=2^2\cdot 3^2\cdot 5\) Thus \(a=2^2\cdot 3\cdot 5~\&~b=3^2\cdot 5\)

 

[Dr.Peterson]

but here i have been given lcm and hcf not two original nos but it happens to be lcm and hcf are multiples of each other so thats why trivially 15 and 180 belongs to set of nos whose lcm and hcf is 15 and 180

Are you saying "but" as an objection, or are you just agreeing with me? I'm not sure whether I have to respond.

The LCM is always a multiple of the HCF, because it is a multiple of both numbers, which are both multiples of the HCF.

 

[Saumyojit]

i am saying why or how 15 and 180 trivially the first set of nos whose lcm is 180 and hcf is 15

 

[Saumyojit]

am saying why or how 15 and 180 trivially the first set of nos whose lcm is 180 and hcf is 15

see i assume that two nos are x and y

x and y will both contain hcf 15 as their factor so prime factorization of x and y will look like
x=15 * non common factors=15*a
y=15 * non common factors =15*b

now from factors list of a and b i have to make sure that i produce both x and y in the factor list of lcm by multiplying and using the least no of factors with hcf being 15.

Eg : SUPPOSE 2 nos 3 and 6 . Factors of 6 are 3*2 & factors of 3 is 3*1.

I know that LCM must contain 3 and 6 in its factors as lcm is divisble by both 3 and 6 . so lcm will first take factor of 3 i.e 3 it deposits 3 in this factor list of lcm {3*......} then it checks how it can produce 6 in its factor list .

IT need a 2 to produce 6 so factor list of lcm borrows only 2 from factor list of 6 {2,3}. now lcm = {3*2} with no extra factors

1. lcm will contain both the factors of x and y with minimum usage of factors.
   in this case lcm will contain =15 * some factor/s from non common factor list {a} * some factor/s from non common factor list {b}so it can produce x and y here in this factor list of lcm.
   
   the equation formed will be like 180=15*a*b where either (15 *a) must give x or y or (15 *b) must give either x or y .
   But we dont know a and b neither x and y

 

[JeffM]

Suppose a, b, c, and d are positive whole numbers and that a * b = c

The smallest multiple of c is c * 1 = c.

If d < b then a * d < c.

Therefore the least common multiple of a and c is c.

What is the greatest whole number that divides a evenly. Obviously a itself. And by hypothesis a divides c evenly.

So the highest common factor is a.

Therefore if a divides c evenly, a is the highest common factor and c is the lowest common multiple

We do not need to bother with prime factorization at all. Just grasp what highest and lowest mean in context.

 

[Dr.Peterson]

As JeffM said, what I called the trivial solution is in fact trivial -- you don't have to do any work. Since 180 is a multiple of 15, you know immediately that their HCF is 15, the smaller number, and their LCM is 180, the larger number.

But I want to show you a nice way to organize your thinking about the factors, in terms of a Venn diagram. Here is the trivial solution:


The numbers inside each ring are the prime factors we multiply to get that number. The HCF is the product of the factors the share (in the overlap), while the LCM is the product of all the factors, taken once each. (No factor can appear on both sides, because that would be a common factor and would have to be put in the middle.)

Here, I put the 3 and 5 in the middle as I had to, and I put all the rest of the factors of 180, the LCM, in the right part. So the numbers x and y are 15 and 180.

Here is the only other way I can arrange them following the rules:


That's the non-trivial solution.

By the way, this also provides a nice way to see that LCM*HCF = xy, because the product xy consists of all the factors, with those in the HCF taken twice; and the same is true of LCM*HCF.

 

[Steven G]

Factored: \(15=3\cdot 5~\&~180=2^2\cdot 3^2\cdot 5\) Thus \(a=2^2\cdot 3\cdot 5~\&~b=3^2\cdot 5\)

???

 

[pka]

Factored: \(15=3\cdot 5~\&~180=2^2\cdot 3^2\cdot 5\) Thus \(a=2^2\cdot 3\cdot 5~\&~b=3^2\cdot 5\)

???

To find the \(GCF\) factor the two numbers. List all common factors and use the least power of the factors.
To find the \(LCM\) factor the two numbers. List all factors in either and use the greatest power of the factors.
Using those two then we find possible \(a~\&~b\) so \(GCF(a,b)=15~\&~LCM(a,b)=180\)

 

[Steven G]

To find the \(GCF\) factor the two numbers. List all common factors and use the least power of the factors.
To find the \(LCM\) factor the two numbers. List all factors in either and use the greatest power of the factors.
Using those two then we find possible \(a~\&~b\) so \(GCF(a,b)=15~\&~LCM(a,b)=180\)

Exactly, but did you do as you said?

 

[pka]

Exactly, but did you do as you said?

Of course not! This is a learning site. So why should we?
That is why we should leave the reader hanging. You should read more on the Moore Method of mathematics education.

 

[Saumyojit]

yes it intutively makes sense hcf * lcm = product of two nos

hcf *(hcf* non common factors between x and y) = all the factors of two nos multiplied. thanks btw

 

[Saumyojit]

@Dr.Peterson i did it like this way and it is giving the correct sets of nos but one doubt :
lcm is 180 and hcf is 15 . x and y are the possible nos --->
x=15a y=15b a and b are two non common factors .

180=15 * a*b as hcf = lcm * non common factors of both x and y
ab=12 now the possible combination may be a=1 b=12 ,a=2 b=6,a=3 b=4

THe first set of non common factors that satisfies the lcm is (1,12) i.e 180=15*1*12

so puting the values of a and b we will get the first set of nos
x=15 a =15 *1=15
y=15 b =15 *12=180.

So the non common factor of 15 is coming as 1 .(a=1)
but if we break 15 and 180 into their prime factors
15=5*3
180=5*3 *3*2*2 here we can clearly see there is no non common factor of 15 but in the above case it is showing a=1

Thats becoz fundamental theorem of arithmetic do not allow 1 to be as a prime no so i am not getting it in prime factorization list but then when i am considering a=1 in x=15a then i am considering as only normal factor.

 

[Dr.Peterson]

@Dr.Peterson i did it like this way and it is giving the correct sets of nos but one doubt :
lcm is 180 and hcf is 15 . x and y are the possible nos --->
x=15a y=15b a and b are two non common factors .

180=15 * a*b as hcf = lcm * non common factors of both x and y
ab=12 now the possible combination may be a=1 b=12 ,a=2 b=6,a=3 b=4

THe first set of non common factors that satisfies the lcm is (1,12) i.e 180=15*1*12

so puting the values of a and b we will get the first set of nos
x=15 a =15 *1=15
y=15 b =15 *12=180.

So the non common factor of 15 is coming as 1 .(a=1)
but if we break 15 and 180 into their prime factors
15=5*3
180=5*3 *3*2*2 here we can clearly see there is no non common factor of 15 but in the above case it is showing a=1

Thats becoz fundamental theorem of arithmetic do not allow 1 to be as a prime no so i am not getting it in prime factorization list but then when i am considering a=1 in x=15a then i am considering as only normal factor.

I'm not sure what you are saying. What does "non common factor of 15" mean to you? In this example, 15 is the greatest COMMON factor, not an uncommon factor!

I suppose what you mean is that you don't explicitly see 3*5 times something else in the prime factorization, so you would not have noticed this as a possible answer. That may be why I initially missed it myself! But I think you see why that doesn't mean that a=1 should be ignored.

Possibly what you mean by "non common factor" is "coefficient", which means "the other factor", as when we say the coefficient of x in 4x is 4. We usually use it to mean the numerical coefficient as in that simple example, but we can also say that the coefficient of x^2 in 15x^3y^2 is 15xy^2. So here you could say that the coefficient of 15 in the factorization of 15 is 1. I have discussed this at length here.

What is your point? Have you answered your own question, or not?

 

[Saumyojit]

I'm not sure what you are saying. What does "non common factor of 15" mean to you? In this example, 15 is the greatest COMMON factor, not an uncommon factor!

Non common factor means which does not fall in the hcf region of two nos . 180 has its hcf 15 and its own factor b as 12 . 12 is the factor that is not common . as 180 is a exact multiple of 15 so all of the factors of 15 are used up and there is no other factor of 15 left that is not used up in hcf region.

A simple eg : 24=2*2*2*3 32=2*2*2*2*2
so 24=8 *3 32=8*4
so 24 has its own non common factor 3 and 32 has its own non common factor 2*2

note that if i represent 24 as =8a and 32 as =8b and if u draw a venn diagram then 8 would be in the overlapping position and at one side there would be 3 and the other side 2*2

NOW in this case all the factors of 15 are common and are in the hcf region there is no factor left . So my intution says a should be zero as a represents non common factor of x ; there is no factor left to be placed in a but that cannot be as x will become zero.

x=15 a
y=15 b so a and b are non common factor" or "the other factor which does not fall in the hcf region .

as shown above ab=12 now the first possible combination of a=1 b=12 will be substitued in x=15a and 15b

as a is coming 1 it should be like x1=15 (common factor) *a ------> x1= 15 *1 so x1 is 15 having common factors (5 *3) and the other non common factor as 1 as a defines non common but 1 is common to every other no so 1 does not fit into the defination of a

 

[Dr.Peterson]

Non common factor means which does not fall in the hcf region of two nos . 180 has its hcf 15 and its own factor b as 12 . 12 is the factor that is not common . as 180 is a exact multiple of 15 so all of the factors of 15 are used up and there is no other factor of 15 left that is not used up in hcf region.

NOW in this case all the factors of 15 are common and are in the hcf region there is no factor left . So my intution says a should be zero as a represents non common factor of x ; there is no factor left to be placed in a but that cannot be as x will become zero.

x=15 a
y=15 b so a and b are non common factor" or "the other factor which does not fall in the hcf region .

as shown above ab=12 now the first possible combination of a=1 b=12 will be substitued in x=15a and 15b

as a is coming 1 it should be like x1=15 (common factor) *a ------> x1= 15 *1 so x1 is 15 having common factors (5 *3) and the other non common factor as 1 as a defines non common but 1 is common to every other no so 1 does not fit into the defination of a

Look back at my post #9. I showed all this! The first picture is of the case you are asking about, where there are no prime factors in the left-hand region. That represents, not 0, but 1: when you don't multiply by anything, it's the same as multiplying by 1. This is the same idea as the 1 you are left with when you cancel all the factors of the numerator of a fraction, like 2/6 becoming 1/3, because you divide both 2 and 6 by their common factor, 2.

And you seem to be correctly recognizing "a" as being 1. That's why I asked whether you had answered your own question.

Now in the last line here you finally show what's troubling you. Your made-up term "non-common factor", which I would call "the other factor" (or the coefficient), can indeed be 1; there is nothing you said in defining what you mean by it that is contrary to that! You seem to be imagining an unstated assumption that a and b have to be products of primes. They simply have to be relatively prime numbers. That is, they can't have any common factors between them other than 1. And 1 has no factors other than 1 in common with 12, so these are perfectly good.

The standard terms we use have been carefully defined to avoid confusion; you are using some in ways that are confusing you. In part, I think you are confusing "common factors" with "common prime factors"; and by not using the term "relatively prime", which explicitly means "having no common factors other than 1", and using your term "non-common" which seems to mix together my "other factor besides the HCF" with "having no common factors" probably contributes to the confusion.

So let me restate what we are doing here, more carefully. Given HCF(x,y) = 15 and LCM(x,y) = 180, we let x = 15a and y=15b, where a and b must be relatively prime. (If they were not, then the LCM would not be 15.) Since LCM(15a, 15b) = 15ab, we know that ab = 12. The only ways to factor 12 into two relatively prime factors are (1)(12) and (3)(4). These lead to the two solutions.

Note that I never mentioned primes here. They are useful in my Venn diagram approach, but seem to be getting in your way for the algebraic approach. What I've said here is all that needs to be said. I also didn't need to define a term "non-common factors", which also gets in your way.

 

[Saumyojit]

You seem to be imagining an unstated assumption that a and b have to be products of primes

may be thats why i was thinking a cannot be 1 as 1 is not a prime no .

 

[Saumyojit]

a subscript p and b subscript p I know it represents each exponent of the each factors but how does the min and max of a subscript p and b subscript p process works.(4,6) take this eg
Why they're adding max of X,y and min of x,y?