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:
216 comments Page 12 of 22.

Asif said:   1 decade ago
Please if you know this answer you tell with clarification.

Jayanta atkari said:   1 decade ago
(91 - 43 = 48) = 4*12.

(183 - 91 = 92) = 4*23.

(183 - 43 = 140) = 4*35.

Thus HCF is = 4.

Saumya tyagi said:   1 decade ago
Well I also cannot understand the correct logic of solving this question please explain.

Abhishek Ray said:   1 decade ago
Kindly revert back with logically, how this problem was processed?

VIGNESH KUMAR said:   1 decade ago
Simple way to calculate HCF:

2|48 92 140
-----------
2|24 46 70
-----------
|12 23 35
------------

Stop if you can't divide by common divisor, so answer = 2*2 = 4.

Mandeep said:   1 decade ago
@Prashant.

P is not a dividend its divisor.

Vamshi said:   1 decade ago
48, 92 and 140 = 4 how now am confusing, please explain.

Dhananjay said:   1 decade ago
@ Sanddep:

Have a look at explanation given by @Prasant. We have to solve equation for 'p'. Hence,we eliminate 'r' by taking differences of numbers.

Dove said:   1 decade ago
I don't understand why you write 2*2 and say the answer is 4.

Neeraj said:   1 decade ago
Why we take HCF? When one should take LCM and HCF.


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