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

The given numbers are 43, 91, and 183.

Subtract smallest number from both the highest numbers.
we have three cases:
183 > 43; 183 > 91 and 91 > 43
183 - 43 = 140
183 - 91 = 92 and
91 - 43 = 48.

Now, we have three new numbers: 140, 48, and 92.
HCF of 140, 48, and 92 using the prime factorization method, we get;
140 = 2 x 2 x 5 x 7
48 = 2 x 2 x 2 x 2 x 3
92 = 2 x 2 x 23.
HCF = 2 x 2 = 4.
The highest number that divides 183, 91, and 43 and leaves the same remainder is 4.

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.

4 * 10 = 40 +3.
4 * 22 = 88 + 3.
4 * 45 = 180 + 3.
Remainder is same 3.
Hence number is 4.

Thanks all for the detailed explanation.

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.

Here is the reason why you take HCF of differences:

43 = p * q1 + r;
91 = p * q2 + r;
183 = p * q3 + r;

Here p is the dividend that we need to maximize. Let's solve these linear equations. First, lets move r to one side, since the remainder is constant.

43 - pq1 = r ----> Eq1
91 - pq2 = r ----> Eq2
193 - pq3 = r ----> Eq3

Solving the three equations:

p(q2 - q1) = 91 - 43 = 48.
p(q3 - q2) = 183 - 91 = 92.
p(q3 - q1) = 183 - 43 = 140.

We can rewrite this as:
pK1 = 48.
pK2 = 92.
pK3 = 140.
or,
48 = pK1 + 0.
92 = pK2 + 0.
140 = pK3 + 0.

Here K1, K2, K3 are new quotients. Our goal is still to maximize the value of p. So of course we will take HCF of 48, 92 and 140 to get the maximum p.

Best Answer, well said, Thanks @Logesh.

@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.

Subtract the smallest number from both the highest numbers.
We have three cases:
183 > 43; 183 > 91 and 91 > 43.
183 - 43 = 140,
183 - 91 = 92 and,
91 - 43 = 48.
Why? Please explain.

In the question, this line is mentioned that "find the greatest number which leaves the same remainder in each case".
When u divide all given numbers by option A;
Then you will get the same remainder "3".
So, that the correct answer is "4".