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

We can represent any integer number in the form of: pq+r.
Where p is dividend, q is quotient, r is reminder.

so: 43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;

Take r same in above three equations as given in 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;

For the greatest value of p that divide each equation we take the HCF of 48,92,140

THEREFORE ANSWER IS 4.

Hey friends I will say one example of HCF.

Eg..72and 126...then,

Using Division method First find H.C.F. of 72 and 126.

72|126|1      
      72
     ----            
      54| 72|1     
            54         
           -----
              18| 54| 3            
                    54               
                   -----
                      0  

H.C.F. of 72 and 126 = 18.

Similarly calculate H.C.F. of 18 and 270 as 18.

Hence H.C.F. of the given three numbers = 18.

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.

HCF of any two number is the number which divide each of them exactly. And any no which divides each of two number exactly also divides their difference, their sum and sum and difference of any multiple of that numbers.

Ex- if 5 is them common factor of 25 and 15. It is also a factor of (25-15) , (25+15) and (25a+15b).

Therefore here we have taken the difference to find the HCF. Because when we will subtract them remainder will get cancelled.

The smallest number divisible by 12,32,42 and 63 is its LCM which is 2016.

The prime factorization of 2016 is 2x2x2x2x2x3x3x7.

Pick out the pairs in this step. They are 2x2 2x2 3x3.

2 and 7 are not in pairs. But if there is one more 2 and one more 7, both 2 and 7 can also be in pairs.

So the required numbers are one more 2 and one more 7. So it is 2x7= 14. Therefore, 2016x14 = 28224.

So the required square number is 28224.

Thank you!

Better to go by basic methods and simple operations to arrive the answer. So taking the options and dividing each no is the best solution. Don't complicate things by doing too much of research.

Just solve it in a smart and easy way and arrive the answer. Wherever the concept is actually required use it only in those cases and not in all the cases. Because the smart and easy approach is more important than the steps.

U guys are not understanding the meaning of H.C.F actually...i'll give an example and an explanation!
Highest common factor
Here are the list of prime factors of 24 and 36:
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
If we write down the numbers that are the same in both lists, they will give us the highest common factor of 24 and 36:
HCF of 24 and 36 is 2 x 2 x 3 = 12
This is the meaning of H.C.F...so simple! :D

Let x be the greatest possible number such that it leaves the same remainder when it divides 183, 91 or 43.

Since the remainder is the same in each case, the difference of the terms must be exactly divisible by x. Also, x must the greatest possible number that exactly divides the difference between the terms.

Required number, x = HCF of (183 " 91, 91 " 43, 183 " 43) = HCF of (92, 48, 140) = 4.

Hi.

In question, they gave a greatest number, in the sense.

Ex- if you take 4 as a greater num for the given question.

Means this is the greater num which divides all the dividend if you take 5 it won't divide all' the dividends.

If they ask greater num -then take smaller num in the choices and divide all the dividends by that num if you get remainder same then that is the answer.

let the greatest number is H.

Now;

H*Quotient(Q1)+Reminder(R)= 43 -------> (1)
H*Quotient(Q2)+Reminder(R)= 91 -------> (2)
H*Quotient(Q3)+Reminder(R)= 183 -------> (3)
Now, (1)-(2), (3)-(2), (3)-(1)
H*(Q2-Q1)= 48,
H*(Q3-Q2)= 92,
H*(Q3-Q1)= 140.

So 48, 92 and 140 are completely divisible by H, that will be the HCF of these numbers.
So, the HCF of 48, 92 and 140 is 4.