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

Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 1)
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
4
7
9
13
Answer: Option
Explanation:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

     = H.C.F. of 48, 92 and 140 = 4.

Discussion:
210 comments Page 17 of 21.

Kiran Kumar said:   9 years ago
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.

Senthmaizhan said:   9 years ago
All this explanation are not clear to me. Please explain it in a simple way.

SHYAM YADAV said:   9 years ago
What are the numbers to find the difference?

Mr.newton said:   9 years ago
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?

Ajeeth said:   9 years ago
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?

Pradeep said:   9 years ago
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.

Brianna said:   9 years ago
I don't understand I'm only eleven, please someone help me to get this.

Khushi said:   9 years ago
I'm confused by this one, Please explain with division method.

Vimalsankar said:   9 years ago
43/4 the remainder is 3. 91/4 the remainder is 3. 183/4 the remainder is 3. So, finally we get the answer 4.

Phani said:   9 years ago
Is this the common HCF problem if so when I try to solve it using Euclid's method I got the answer '1'.

Am I right?


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