# 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.

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:

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.

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)

Madhu said:
9 months ago

Take difference:

43 - 91 = 48.

91 - 183 = 92,

Now take the hcf of difference numbers 49, 92.

The HCF Of 48, 92 = 40.

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.

@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.

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.

(Divisor x quotient) + remainder= dividend.

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