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:
212 comments Page 5 of 22.

Hary said:   2 decades ago
Solving the equations:
91=pq2+r
43=pq1+r
_________________
48=pq2-pq1
thus
48=p(q2-q1)=91-43.

Shwetha said:   2 decades ago
Simply divide each number by 4 the remainder is same for all so the answer is 4.

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.

Nam said:   1 decade ago
Acc to the formulae H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number.

So as per the first formulae the hcf of 183 & 43 is comming 10
n the third no. is 91 so how will 4 be the ans

Hind said:   1 decade ago
I go with same method as shweta and dhanasekar has said in the forum.

Navneet said:   1 decade ago
Divide each with each option.

Pankaj maharaj said:   1 decade ago
Yesss! you sud simply divide and find the ans in quicker way.

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

Titus said:   1 decade ago
That is true. Understanding is more important.

Sweety said:   1 decade ago
I don't understand why we need to find the difference. Please explain the logic behind it.


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