Question

take any two numbers.find their lcm and hcf.what is your ovservation​

Answer 1

Step-by-step explanation:

How do I find the LCM when the product and HCF of two numbers is given?

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We use the relation that for any two integers a,b , that:

hcf(a,b)×lcm(a,b)=ab

Rearranging, we get:

lcm(a,b)=abhcf(a,b)

To understand how this relation comes about, we first write the prime decomposition of the two integers:

a=2α1×3α2×5α3×...

b=2β1×3β2×5β3×...

Now, the hcf function can be written as follows:

hcf(a,b)=2min(α1,β1)×3min(α2,β2)×5min(α3,β3)×...

Why does it take this form?

Consider a=16=24 and b=64=26 . The HCF is given as hcf(16,64)=2min(4,6)=24=16 . The prime factors of the HCF can only have an exponent as large as the smaller exponent between the two numbers.

Similarly, the lcm function can be written as follows:

lcm(a,b)=2max(α1,β1)×3max(α2,β2)×5max(α3,β3)×...

Finally, we multiply them together, to get

hcf(a,b)×lcm(a,b)

=2min(α1,β1)×3min(α2,β2)×5min(α3,β3)×...2max(α1,β1)×3max(α2,β2)×5max(α3,β3)×...

=2min(α1,β1)+max(α1,β1)×3min(α2,β2)+max(α2,β2)×5min(α3,β3)+max(α3,β3)×...

We then make the simple observation that min(α,β)+max(α,β)=α+β , and our expression simplifies to

hcf(a,b)×lcm(a,b)

=2α1+β1×3α2+β2×5α3+β3×...

=(2α1×3α2×5α3×...)(2β1×3β2×5β3×...)=ab , as desired.

Answer 2

$\underline\mathfrak{ANSWER:-}$

If the two numbers are aa and bb and their LCM is xx and their HCF is yy then there is a rule that is a×b=x×ya×b=x×y

Let's take an example. HCF of two numbers is 3 and LCM of those numbers is 90.

Now product of HCF and LCM = 270. Is there anyway to find out those numbers? Now find the factors of 270 and verify Y is for yes and N is for no

2×135LCM(N)HCF(N)2×135LCM(N)HCF(N)

3×90LCM(Y)HCF(Y)3×90LCM(Y)HCF(Y)

5×54LCM(N)HCF(N)5×54LCM(N)HCF(N)

6×45LCM(Y)HCF(Y)6×45LCM(Y)HCF(Y)

9×30LCM(Y)HCF(Y)9×30LCM(Y)HCF(Y)

10×27LCM(N)HCF(N)10×27LCM(N)HCF(N)

15×18LCM(Y)HCF(Y)15×18LCM(Y)HCF(Y)

Possible pairs of numbers are :

(3, 90), (6, 45), (9, 30) and (15, 18).

All the above mentioned pairs of numbers have 3 as HCF and 90 as LCM.

If LCM and HCF are given then there can be many possible pairs of numbers that fulfill the given criteria.