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:
216 comments Page 2 of 22.

Priti Maurya said:   2 years ago
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".
(43)

Abrar said:   2 years ago
By division method:

43)183(4
172
-----
11)43(3
33
------
10)11(1
10
-----
1)10(10
10
------
0
Now use the last divisor with the mid-number 1.
You get again the remainder as a 0,
Hence 1 should be the answer.
(22)

Anurag Parashar Sarmah said:   2 years ago
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.
(68)

Mayur Jagtap said:   2 years ago
4 * 10 = 40 +3.
4 * 22 = 88 + 3.
4 * 45 = 180 + 3.
Remainder is same 3.
Hence number is 4.
(88)

Nirmod Kumar Mandal said:   2 years ago
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.
(31)

Sri Sekhar said:   3 years ago
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.
(308)

Pavi said:   3 years ago
Thanks all for the detailed explanation.
(63)

Piyush said:   3 years ago
Best Answer, well said, Thanks @Logesh.
(38)

Sanjana said:   3 years 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.
(53)

Madhu said:   4 years ago
Take difference:

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

Now take the hcf of difference numbers 49, 92.

The HCF Of 48, 92 = 40.
(19)


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