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

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
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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

THEREFORE ANSWER IS 4.

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.

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?

I also want an explanation about the logic of the solution.

Explain the logic of this solution.

Why they took difference between those nos.

Please explain me.