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 9 of 21.

Samir said:   1 decade ago
Which is the highest five digit number which is equally divisible by 60, 80, and 90 and how to arrive this answer ?

Shital said:   1 decade ago
@Parthiban.

If there is large numbers then it is reliable method to divide and then finding highest common factor?

RAVI VARMA 501 said:   1 decade ago
Swetha and prashanth explanation is good. And abve some are all in kids formulas. . . So dont look others friends.

Sarath said:   1 decade ago
I think the smarter way will be to divide the numbers with options and check. But your speed has to be so high.

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.

Sakthivel said:   1 decade ago
How to arrive at this equation p(q2-q1)= (91-43) from 43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;

Dhanasekar said:   1 decade ago
Simply try divide to divide each by the options, by this which is exactly divided then that is the answer.

Durga said:   1 decade ago
Getting answer is not important but knowing how to approach that answer is important. Thanks for Prashant.

Pabitra Bhusan Saha said:   7 years ago
Your explanation is very easy to understand and this is the best process, Thanks @Saroja and @Sourav.

RDbaby said:   1 decade ago
@ Merry.

Pratibhan meant, the common thing coming in prime factor list of all 3 nos is 2*2, i.e 4.


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