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.
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:
213 comments Page 5 of 22.
Sakthivel said:
2 decades ago
How to arrive at this equation p(q2-q1)= (91-43) from 43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;
91 = pq2 + r;
183 = pq3 + r;
Hary said:
2 decades ago
Solving the equations:
91=pq2+r
43=pq1+r
_________________
48=pq2-pq1
thus
48=p(q2-q1)=91-43.
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:
2 decades ago
Simply try divide to divide each by the options, by this which is exactly divided then that is the answer.
Nam said:
2 decades 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
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:
2 decades ago
I go with same method as shweta and dhanasekar has said in the forum.
Navneet said:
2 decades ago
Divide each with each option.
Pankaj maharaj said:
2 decades ago
Yesss! you sud simply divide and find the ans in quicker way.
Durga said:
2 decades ago
Getting answer is not important but knowing how to approach that answer is important. Thanks for Prashant.
Titus said:
2 decades ago
That is true. Understanding is more important.
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