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
Answer: Option
Explanation:

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

     = H.C.F. of 48, 92 and 140 = 4.

Discussion:
210 comments Page 3 of 21.

Kuldeep Sharma said:   7 years ago
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.
(3)

Ram rao said:   3 years ago
Thanks everyone for explaining the answer.
(3)

Raj said:   2 decades ago
Why do we do the difference of number to each other?
(1)

Govind Paswan said:   1 decade ago
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.
(1)

Tamzid Khan said:   9 years ago
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.
(1)

Logesh said:   8 years ago
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.
(1)

Shreekanth said:   7 years ago
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.
(1)

Bala said:   6 years ago
@All.

They asked to divide by a number also we need to have the same remainder in all cases.

So, dividing by 4, we get remainder 3 in all cases.
(1)

Mozammil anwar said:   6 years ago
4/43 = remainder is 3.
4/183 =remainder is 3.
4/91 = remainder is 3.
So that' why the answer is 4.
(1)

Vamsi said:   5 years ago
48 = 4 * 12
92 = 4 * 23
140 = 4 * 35.

1,2,3,5 prime numbers so 4 is the correct answer.
(1)


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