Aptitude - Problems on H.C.F and L.C.M - Discussion

There is the difference andthe cut remainder due to following reason ...
@Prashant

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

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

Take r same in the above three equations as given in the 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.

Here, you have to take the difference due to cancel the remainder and cancellation is applied
{
p(q2-q1)= (91-43)=48; (to get the p[divisor*which is common among all of them eq 1,2,3*** ]and two quotient is given.
p(q3-q2)= (183-91)=92;
p(q3-q1)= (183-43)=140; }.

Let neglect two quotient[q2-q1..etc]//// 'p' is that value which had to be common so you have to calculate given HCF of eqns.

48 = 4 * 12
92 = 4 * 23
140 = 4 * 35.
Here consider p is 4 [which is hcf &&common..which is 'p' remember] ..and 12 is quotient [q2-q1].

I hope you will understand.

@All.

According to me,

the solution is;

Factors of 43 are: 1, 43.
Factors of 91 are: 1, 7, 13, 91.
Factors of 180 are: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180.

So, I go with my answer i.e., 1. None of the given options is correct.

I am not understanding this. Please explain in detail.

Thanks everyone for explaining the answer.

Thanks everyone for explaianing.

P is called the divisor & not dividend.

(Divisor x quotient) + remainder= dividend.

Thanks everyone for explaining.

Why they subtracted? Explain, please.

Thanks all for giving the clear explanation.

Why they took difference here? Anyone explain about it.