Aptitude - Problems on H.C.F and L.C.M - Discussion

HCF means highest common factor.

LCM means least common Multiple.

Both are different. In this question they have given HCF of two numbers are 23.

Let us consider x and y as two no's then HCF (x, y) = 23.

We don't know the value of x and y. Then LCM of the two nos have factors 13 and 14.

This means that LCM (x, y) = z.

The value z have factors 13 and 14 which will divide z without leaving a remainder.

The main question is finding the larger of the two no is? That means we have to find biggest value among x and y.

I have understood the question but I don't how to solve it. I need clear explanation please.

Statement 1:- We know that if 2 numbers are co-primes, then their LCM is equal to their product.
Statement 2:- We also know that if 2 numbers are co-primes,then their HCF is 1.

In the given question, 13 and 14 are the factors of the LCM of 2 numbers. Therefore it is understood that the 2 numbers are the multiples of 13 and 14.

23 is the HCF of 2 numbers. Therefore 2 numbers are not co-primes.
Given that 23 is the common factor of the 2 numbers - 23x13 = 299 and 23 x 14 = 322.
LCM of 299 and 322 = 4186.13 and 14 are the factors of 4186.
HCF of 299 and 322 = 23.
Therefore, 23 x 14 = 322 is the highest number.

Frnds Try to solve it in reverse. You will understand what the question is.
Means find HCF and LCM of,

23*13=299 and 23*14=322.
Factors of 299 =13*(23).
Factors of 322=2*7*(23).
So HCF is 23 that is given in the question.

Next thing in question is other two factors of their LCM are 13 and 14. To understand this look at the factors of these two numbers (299 & 322).

13 is there in factors of 299 and 14 (2*7) is present in factors of 322.

This explanation will help you to understand what the question is saying. If you understood the question u can easily solve other questions like this.

Guys, Let's understand it by set theory. We have set A={a,b,c,d} & B={e,b,c,f,g}.and in this question abcd are factors of A and ebcfg are factors of B.

So hcf will be b.c and that's common in both factors.If we say lcm of A, B then It will contain b,c as common part of their lcm({b,c}, adefg) where ad belongs to A and efg belongs to B and we will call them uncommon or other factors of their lcm.and you know that how are they used to get a particular no.A or B. You can take ad as 13 and efg as 14.

So, lcm will be 23 .13.14 and hcf will be 23 and no.A is 23.13 and B is 23.14.

Here in the question it is given that, " Other two factors are 13 and 14".
It means L.C.M of the number will contain these two along with the given common factor 23.
Thus L.C.M can be:

L.C.M(a,b)=23*13*14;
Now, we know ;
L.C.M(a,b) * H.C.F(a,b)= a*b;
=> ( 23*13*14)*(23)=a*b; ----> 1
Now, as 23 is the common factor so both number must be a multiple of 23.
So, Let us say numbers are 23*x, and 23*y.
Now, a*b =23*x * 23*y ----> (ii)
Therefore, from (i) and (ii).
We can conclude, a=23*13 and that b=23* 14. Or, vice-versa.
And, obviously greater among them will be 23*14.

Let 'a' and 'b' be the two numbers.

HCF (a, b) = 23.

LCM (a, b) = 13*a = 14*b (or) LCM = 14*a = 13*b.

Note : The (or) case is just to indicate that we could have taken the factors the other way round ---> since they are factors for LCM. Both cases would result in the same large number.

Theorem: The product of two numbers = a*b = LCM (a, b)*HCF (a, b).

Applying the above theorem here :

=> a*b = HCF (a, b)*LCM (a, b).

=> a*b = 23*(13*a) (or).

Therefore => b = 23*13.

Similarly, a = 23*14.

The largest of the 2 numbers = 23*14 = 322 is the answer.

Perhaps this answer is not correct:

As we see HCF= 23.
LCM= 23*13*14.
Let the numbers be a and b.
HCF*LCM=a*b.
a*b=23*23*13*2*7.

So we have to take 23 as common in each number as it is HCF.
So a=23*m,
b=23*n,
now m*n=13*2*7.

As HCF is 23 so m and n should be co prime.
Hence the possible pairs of m and n = (13,14), (26,7), (91,2).
As the value of 91 will give the highest value so we have to select 91,2 as a solution.
m=91 n=2.
and a=2093 and b=46.

Verify yourself LCM of 2093 and 46 is 23*13*14 and HCF of it is 23.
So the biggest possible number is 2093.

1. We know that the HCF of a prime number (when it same i.e. 23 and 23) so the HCF of 23, and 23 is 23 ok.

2. Then in question there is no LCM but here we know that the LCM will be the same as HCF when we are talking about the LCM of 23 and 23. So LCM is 23.

3. But there is a problem that 23 can not be divided by 13 and 14, hence they are not the factors of 23.

But I say they are not the factors but we make 13 and 14 factors of 23 by multiplying them with 23 ok so let's do it. 23'*13=299, 23'-14=322. Then check 13 and 14 completely divided 299 and 322.

HCF of 2 numbers equal to the highest of all the common factors of the two numbers
23 is the common factor of the two numbers.

Let 23 x,23 Y be the two numbers.
we know that the factor of any two numbers is also a factor of the LCM of those two numbers.
So 23 is a factor of the two numbers 23 x and 23 y.
LCM of two numbers.
23 x and 23 Y = 13* 14 *23.

LCM =13*14 *23,
And HCF = 23,
We know that,
LCM * HCF = product of the two numbers,
13 *14 * 23 =23 x * 23 y.
@x*y= 13* 14.

Which shows that larger number will be 14 * 23 that is equal to 322.

When we calculate hcf of two numbers we find factors which are common.

And for lcm we take common factor and the remaining factors of two.

Here it is given that the other factors of lcm are 13 and 14

So numbers are hcf*lcm part

=>23*13 and 23*14 so 23*14 is greater.

Example: suppose we have two numbers 12 and 15.
Factor of 12=2*2*3.
Factor 0f 15=3*5.
So hcf =common factor=3.
And lcm=2*2*3*5=60 .
lcm part of 12=2*2.
lcm part of 15=5.
If we multiply these lcm parts by hcf we get the corresponding numbers.