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

1st you have to subtract them all by each higher no.
(183-43) = 140.
(183-91) = 92.
(91-43) = 48.
Now find the common factor of these new numbers.
i.e 4.

So the correct answer is option A.

Assume that the required answer is "D" and the remainder that it leaves when dividing 43, 91 and 183.

This means that,

43 = d*q1 + r,

91 = d*q2 + r,

183 = d*q3 + r,

Where q1, q2 and q3 are the respective quotients.

Subtracting the first equation from second, we get:

91 - 43 = (d*q2 + r) - (d*q1 + r).

Hence, 48 = d*(q2 - q1) = d*(an integer).

Similarly, subtracting second equation from first, we get:

183 - 91 = (d*q3 + r) - (d*q2 + r).

Hence, 92 = d*(q3 - q2) = d*(another integer).

Now, 48 = d*(an integer) and 92 = d*(another integer) mean that both 48 and 92 are divisible by d. (In other words, "d" is a common divisor of 48 and 92. )

Now common divisors of 48 and 92 are 1, 2 and 4 only. Out of which, highest is 4. (Alternatively, you can directly find the HCF (or GCD) of 48 and 92). Hence 4 is the answer.

Note:

There is no need to calculate the HCF of 3 numbers, namely (91 - 43), (183 - 91) and (183 - 43). It is sufficient to take any two of these three numbers. The HCF will still be 4.

Thanks everyone for explaianing.

let the greatest number is H.

Now;

H*Quotient(Q1)+Reminder(R)= 43 -------> (1)
H*Quotient(Q2)+Reminder(R)= 91 -------> (2)
H*Quotient(Q3)+Reminder(R)= 183 -------> (3)
Now, (1)-(2), (3)-(2), (3)-(1)
H*(Q2-Q1)= 48,
H*(Q3-Q2)= 92,
H*(Q3-Q1)= 140.

So 48, 92 and 140 are completely divisible by H, that will be the HCF of these numbers.
So, the HCF of 48, 92 and 140 is 4.

Thanks everyone for explaining the answer.

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

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

Explanation: 43, 91, 183.
183 = (183-91) = 92.
91 = (183-43) = 140.
43 = (91-43) = 48.

92, 140, 48.

92 = 2*2*23.
140 = 2*2*35.
48 = 2*2*2*3*2.

Ans 4.

Is there anyone who will clearly explain why one has to take the difference of those numbers?

I mean what is the key note in the question seeing which I will solve the problem in this way?

Please explain someone.

The answer is 4 and how it is 4 is below,

We can represent any integer number in the form of: D*q + r.
Where D is divisor, q is quotient, r is the remainder.

So each number can be written accordingly:
43 = D*q1 + r1;
91 = D*q2 + r2;
183 = D*q3 + r3;

r1, r2 & r3 will be same in above three equations according to the question.
D is the value that we want to find out. which should be greatest.

On solving three equations we get:

D*(q2-q1)= (91-43)=48
D*(q3-q2)= (183-91)=92
D*(q3-q1)= (183-43)=140
It is obvious that q3>q2>q1.

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

THEREFORE ANSWER IS 4.

Keep on dividing all the three numbers by 2 so that reminder should be the same ie
2[43 91 183]
[ 21 45 91] remainder will be 1 for all the three.

Divide again by 2.
2[21 45 91]
10 22 45 remainders will be same ie 1.

So after this, if you try to divide all three again by 2 then you get a different value for the reminder.
And hence; 2 * 2 = 4.