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

Dear friends, I am trying to satisfy this question, that why we have to take differences of those numbers.

We all know that the result of H.C.F of any two or more numbers is only their difference between them, such as H.C.F. OF 7& 14 = 7.

Now we calculate the H.C.F. of 7, 14 &21.
We see that the differences between them are actually (14 - 7 = 7, 21 - 7 = 7 & 21 - 7=14.

But the difference of 21 - 7 = 14, whose factor is 2 & 7.

Hence, only 7 will be the H.C.F. OF 7, 14 & 21.
On the above basis, the difference between them is 48, 92 & 140.

Therefore, the H.C.F.of 48, 92 & 140 will be 4.

Hi, how to solve the Lcm? Please help me.

Better to go by basic methods and simple operations to arrive the answer. So taking the options and dividing each no is the best solution. Don't complicate things by doing too much of research.

Just solve it in a smart and easy way and arrive the answer. Wherever the concept is actually required use it only in those cases and not in all the cases. Because the smart and easy approach is more important than the steps.

All this explanation are not clear to me. Please explain it in a simple way.

What are the numbers to find the difference?

Everything is fine, but I am getting doubt with HCF. In three numbers common we get 2. If we divided 2 * 2 * 2 get 8. Why are you telling 4?

How to take HCF I don't know how to take HCF please anybody explains me. How to find the HCF for any unknowns for ex: 533, 522, 455 for this number?

Hi @Ajeeth, very simple for the H.C.F.

For the numbers, you have given (455, 522 & 533).

H.C.F=1, Because the factor of these numbers are 455 = 1 x 5 x 7 x 13,
522 = 1 x 2 x 261 &
533 = 1 x 533.

So except 1, no digit is common in each number. Hence the common digit 1 is the H.C.F of above numbers.
This is the common rule of finding H.C.F. That those digit/digits which will be common in all the numbers.

Eg. You have to find the H.C.F. of 24, 60 & 81. Now factors of each number will be 24 = 2 x 2 x 2 x 3, 60 = 2 x 2 x 3 x 5, & 81 = 3 x 3 x 3 x 3.

Here only the digit '3' is common in all, hence 3 will be H.C.F of these numbers. Also note that if more than 1 digit is common in all the numbers then multiplication of those digits will be the H.C.F (suppose 2 & 5 both are common in all the numbers then the multiplication of both the digit will be 10) hence 10 will be the H.C.F of those numbers.

Thanks.

I don't understand I'm only eleven, please someone help me to get this.

I'm confused by this one, Please explain with division method.