LCM and HCF

This chapter deals with finding the lowest common multiple and Highest Common factor for n numbers.

Importance: This chapter although simple regularly is selected by SSC in the exams and has a weightage of about 2 questions per 100. I would frequently add the questions and thus it would be a nice and complete compendium for preparation.

HCF

This refers to the highest common factor that is actually the common factor of n no. and the highest amongst them so that the numbers are completely divisible by them. Also known as Greatest Common Divisor.

The method explained above is prime factorization method.However, there is another method that is division method


8|12| 1
    8
------
     4|  8 |2
          8
        ----
          0

Thus 4 is the divisor that completely divides both 8 and 12 and thus is the highest common factor.


This analysis however shows one more thing the subtraction of the two no. i.e. 12 and 8 is 4 i.e the HCF. Thus it can be deduced that subtraction of the two no. would  give a multiple of the HCF or the HCF itself.

Now how to determine HCF quickly for three no.

Find the H.C.F. of 72, 126 and 270.

Solution

Using Prime factorisation method
    72 = 2×2×2×3×3 = 2³×3²
  126 = 2×3×3×7 = 2¹×3²×7¹
  270 = 2×3×3×3×5 = 2¹×3³×5¹
H.C.F. of the given numbers = the product of common factors with least index
  = 2¹×3² = 2×3×3 = 18

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

Method of difference for above question

No. are 72 126 and 270
now 126-72= 54 and 270-126=144 and 270-72=198
consider the lowest 2 that is 54 and 144. Identify the HCF for them by subtraction method. Subtraction gives=144-54=90.
This must be the multiple of HCF now submultiple could be 1,2,5,9,10,18 and highest is 18 thus HCF is 18

Practice this method and you would be able to find the HCF quickly

SSC VIEW OF HCF

After this basic view I would move on to the basic questions asked in SSC based on HCF. It is requested to the readers(if any) to please post the questions related to HCF in comments section and i would update them over here so that it would benefit everybody. In this way I would be able to cover your doubts as well as you would contribute to enlarge this compendium.

Q1. What is the greatest number that will divide 2400 and 1810 and leave remainders 6 and 4 respectively?

Ans. Here the no. would have been completely divisible by that greatest number if there had been no remainder. thus the nos. that are completely divisible by the number we have gto find are: 2400-6=2394 and 1810-4=1806.
Now the greatest no. that would divide 2394 and 1806 both would be its HCF.
Thus 2394-1806=588
=1806-3*588=1806-1764=42.

You can also get this by division method.


Practice Question
Q2. What is the greatest number that would divide 38,45,52 ad leave remainders as 2,3, 4 respectively?     [ 6 ]


Q3. Find the greatest number which is such that when 76,151 and 226 are divided by it the remainders are all alike. Also find the common remainder?

Ans. In these type of questions, you must know that subtraction of two numbers give a multiple of HCF. Now let the common remainder for all be k so the numbers 76-k,151-k,256-k are exactly divisible by the required greatest number.

Now as we know, the subtraction of two numbers would give a multiple of hcf
so, 151-k-(76-k)=75,256-k-(151-k)=75,and 256-k-(76-k)=150 are multiples of required number. The HCF of these numbers 75,75 and 150 is 75.
and the common remainder would be 76/75=151/75=226/75=1








Continued..................