# 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
Explanation:

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

= H.C.F. of 48, 92 and 140 = 4.

Discussion:
200 comments Page 1 of 20.

Pavi said:   2 months ago
Thanks all for the detailed explanation.
(11)

Piyush said:   5 months ago
Best Answer, well said, Thanks @Logesh.
(10)

Sanjana said:   6 months ago
The H.C.F is 1.

The prime factorization of 43 is 43 as it is a prime number.
The prime factorization of 91 is 7 and 3.
The prime factorization of 183 is 3 and 61.

There is no common numbers between these 3 numbers. But,1 is a common factor for each and every number. In this case, 1 is the H.CF.
(15)

Take difference:

43 - 91 = 48.
91 - 183 = 92,

Now take the hcf of difference numbers 49, 92.

The HCF Of 48, 92 = 40.
(12)

Shashwat said:   11 months ago
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.
(23)

Subhronil Biswas said:   12 months ago
@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.
(9)

Ayesha said:   1 year ago
I am not understanding this. Please explain in detail.
(10)

Ram rao said:   1 year ago
Thanks everyone for explaining the answer.
(2)

Narendra Singh said:   1 year ago
Thanks everyone for explaianing.
(2)

Anomie said:   2 years ago
P is called the divisor & not dividend.

(Divisor x quotient) + remainder= dividend.