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 15 of 21.

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.

Tamzid Khan said:   9 years ago
Is there anyone who will clearly explain why one has to take the difference of those numbers?

I mean what is the key note in the question seeing which I will solve the problem in this way?

Please explain someone.
(1)

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?

Ramya said:   9 years ago
I'm not understanding the logic. Please someone explain me.

Tanvi said:   9 years ago
I am not understanding please explain me.

Manoj Rmk said:   9 years ago
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Explanation: 43, 91, 183.

183 = (183-91) = 92.
91 = (183-43) = 140.
43 = (91-43) = 48.
92, 140, 48.

92 = 2 * 2 * 23.
140 = 2 * 2 * 35.
48 = 2 * 2 * 2 * 3 * 2.
Ans 4.

Bisht Ashish said:   9 years ago
@Prashant.

In your explanation, 'p' must be a Divisor, not the Dividend.

Arsh said:   8 years ago
Here, the remainder is unknown that is why we have only chance to find their difference.

Sowmya said:   8 years ago
Well explained, thanks @Prashant.

Mani said:   8 years ago
Why can't we think like this, In given four option one is HCF of 43, 91, 183. And leave the same reminder as per the question.

Let me consider 4 is HCF then divide the numbers by four. It shows the same reminder hence 4 is the answer.


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