# 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
Explanation:

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

Discussion:
200 comments Page 20 of 20.

Simply divide each number by 4 the remainder is same for all so the answer is 4.

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

How to arrive at this equation p(q2-q1)= (91-43) from 43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;

We can represent any integer number in the form of: pq+r.
Where p is dividend, q is quotient, r is reminder.

so: 43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;

Take r same in above three equations as given in question.
p is the value that we want to find out. which should be greatest.

On solving three equations we get:

p(q2-q1)= (91-43)=48;
p(q3-q2)= (183-91)=92;
p(q3-q1)= (183-43)=140;

For the greatest value of p that divide each equation we take the HCF of 48,92,140

If three nos as in this case is 43, 91 and 183 are given then to find HCF the shortest solution is take the diff:

1) 91 - 43 = 48 (4*3*7)
2) 183 - 91 = 92 (4*23)
3) 183 - 43 = 140 (4*7*5)

Thus the HCF is 4.

Anirudh rai said:   1 decade ago
Is this the only way to this type of problem?

If yes then would you please explain it.

Why do we do the difference of number to each other?

Sourabh Das said:   1 decade ago
I also want an explanation about the logic of the solution.