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

43, 91, 183 :

Solution by Factorisation:

The factors of 43 are: 1, 43.
The factors of 91 are: 1, 7, 13, 91.
The factors of 183 are: 1, 3, 61, 183.
Then the greatest common factor is 1.

This is good to gain knowledge. Thanks all.

Good explanation, Thank you all.

Good explanation. Thanks all.

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".

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.

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.

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

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.

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.