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

Divide each with each option.

I go with same method as shweta and dhanasekar has said in the forum.

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

Simply try divide to divide each by the options, by this which is exactly divided then that is the answer.

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.