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HCF And LCM

These are very simple and important topics. The trend has been to ask questions from this section quite frequently nowadays. Although the questions asked will not be direct, it will be a indirect question and thus the biggest challenge in an exam for this topic lies in identification of the question and that it is asked from this topic. Solving a problem after identification is easy.

Note: Calculation speed is very important for this topic esp. division with prime nos. like 2,3 ,7 etc for factorization.

Let’s start with very basic concepts and then go to HCF and LCM.

FACTOR

A factor of a given number is every number that divides exactly into that number.

Example
Write down all factors of 10.
10 = 2 x 5, so numbers 2 and 5 are factors of 10.
Also 10 = 10 x 1, so 10 and 1 are factors of 10.
The factors of 10 are 1, 2, 5, and 10.

NOTE: Number 1 and the number itself are always factors of any number.

PRIME AND COMPOSITE NUMBERS

A prime number has exactly 2 factors, the number itself and 1.

In other words, the prime number can be divided only by 1 and by itself.
NOTE: 0 and 1 are not prime numbers.
Example: 5 is a prime number, because the only factors it has are 1 and 5.
The prime numbers less than 20 are 2,3,5,7,11,13,17,19
Example:
Find all prime factors of 30.
Solution:
All the factors of 30 are 30, 15, 10, 6, 5, 3, 2, 1
But only 5, 3 and 2 are prime numbers.
Thefore all prime factors of 30 are 2, 3 and 5.
A composite number has at least one more factor that the number itself or 1.
In fact, all whole numbers that are not prime are composite except for 1 and 0, which are not prime and not composite.
The composite numbers less than 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, and 18.


DIVISIBILITY RULES

The simple divisibility rules will help you to find factors of a number.

The number is divisible by:
• 2 if the last digit is 0, 2, 4, 6, or 8 (example: 12346);
• 3 if the sum of digits in the number are divisible by 3
(example: 1236, because 1+2+3+6 = 12 = 3 x 4);
• 4 if the last 2 digits are divisible by 4
(example: 897544, because 44 = 4 x 11);
• 5 if the last digit is 0 or 5
(example: 178965 or 40980);
• 6 if it is divisible by 2 and 3;
• 8 if the last 3 digits are divisible by 8
(example: 124987080, because 080 = 8 x 10;
• 9 if the sum of digits is divisible by 9
(example: 234612, because 2+3+4+6+1+2 = 18 = 9 x 2);
• 10 if the last digit is 0
(example: 99990 );
• 11 if the difference between sum of digits in odd and sum of digits in even places is 0 or a multiple of 11.
• 12 if the no is divisible by both 3 and 4.
• 19 if the no of tens added to twice the no of ones is divisible by 19.
(Example: 228; no of tens =22; no of ones = 8; the required sum = 22+ 2x8 =38; hence, 228 is divisible by 19)
• 25 if the no formed by the last two digits (ie digits at tens and ones place) is divisible by 25.
(Example: 31243482325; no formed by last two digits =25;
hence 31243482325 is divisible by 25)
• 100 if the last 2 digits are 0
(example 987600);
• 125 if the no formed by last 3 digits is divisible by 125.

(Note: there are some complex divisibility tests for 7, 13 and 17 too but I recommend not to follow them as dividing the no given would be easier and faster than applying those divisibility tests.)


NOTE: If a number is divisible by 2 factors, it is also divisible by the product of these factors.
Example 1: Number 18 is divisible by 2 and 3, so it must be divisible by 2 x 3 = 6.
Example 2: Number 945 is divisible by 9 (why?) and by 5 (why?), so it must be divisible by 9 x 5 = 45. (Can you check it?)

COMMON FACTORS
When two (or more) numbers have the same factor, that factor is called a common factor.

Example
Find all the common factors of 12 and 18.
Factors of 12 are 1, 2, 3, 4, 6, 12.
Factors of 18 are 1, 2, 3, 6, 18.
The common factors of 12 and 18 are 1, 2, 3 and 6.

HIGHEST COMMON FACTOR (H.C.F)

The Highest Common Factor (H.C.F) of two (or more) numbers is the largest number that divides evenly into both numbers. It is also known as G.C.D. (Greatest common divisor)
In other words the H.C.F is the largest of all the common factors.
The common factors or of 12 and 18 are 1, 2, 3 and 6.
The largest common factor is 6, so this is the H.C.F. of 12 and 18.
It is very easy to find a H.C.F. of small numbers, like 6 and 9 (it is 3) or 8 and 4 (it is 4).
The best way is to keep finding the factors of the smaller number, starting from the largest factor. The first factor of the smaller number that is also a factor of the larger number is a H.C.F.

LEAST COMMON MULTIPLE (L.C.M.)

A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24...
The least common multiple (LCM) of two numbers is the smallest number (not zero) that is a multiple of both.

Concept – Prime Factorisation

A prime factorisation of a natural number can be expressed in the exponential form.
For example:
(i) 48 = 2 x 2 x 2x 2 x3 = 24 x 3
(ii) 420 = 2 x 2 x 3 x 5 x 7 = 22 x 3 x 5 x 7

Methods to Find HCF

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

Using Prime factorisation method
72 = 2 x 2 x 2 x 3 x 3 = 2 3 x 32
126 = 2 x 3 x 3 x 7 = 2 1 x 32 x 71
270 = 2 x 3 x 3 x 3 x 5 = 21 x 33 x 51

H.C.F. of the given numbers = the product of common factors with least index = 21 x 32
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

Co-prime numbers: Two natural numbers are called co-prime numbers if they have no common factor other than 1.
in other words, two natural numbers are co-prime if their H.C.F. is 1.
Some examples of co-prime numbers are: 4, 9; 8, 21; 27, 50.

Methods to Find LCM

Method 1 Simply list the multiples of each number (multiply by 2, 3, 4, etc.) then look for the smallest number that appears in each list.

Example: Find the least common multiple for 5, 6, and 15.
Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,...
Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,...
Multiples of 15 are 30, 45, 60, 75, 90,....

Now, when you look at the list of multiples, you can see that 30 is the smallest number that appears in each list.Therefore, the least common multiple of 5, 6 and 15 is 30.

Method 2 To use this method factor each of the numbers into primes. Then for each different prime number in all of the factorizations, do the following...

1. Count the number of times each prime number appears in each of the factorizations.
2. For each prime number, take the largest of these counts.
3. Write down that prime number as many times as you counted for it in step 2.
The least common multiple is the product of all the prime numbers written down.

Example: Find the least common multiple of 5, 6 and 15.
Factor into primes
Prime factorization of 5 is 5
Prime factorization of 6 is 2 x 3
Prime factorization of 15 is 3 x 5

Notice that the different primes are 2, 3 and 5.

Now, we do
Step #1 - Count the number of times each prime number appears in each of the factorizations...
The count of primes in 5 is one 5
The count of primes in 6 is one 2 and one 3
The count of primes in 15 is one 3 and one 5

Step #2 - For each prime number, take the largest of these counts. So we have...
The largest count of 2s is one
The largest count of 3s is one
The largest count of 5s is one

Step #3 - Since we now know the count of each prime number, you simply - write down that prime number as many times as you counted for it in step 2.
Here they are...2, 3, 5

Step #4 - The least common multiple is the product of all the prime numbers written down.
2 x 3 x 5 = 30
Therefore, the least common multiple of 5, 6 and 15 is 30.

Relation between L.C.M. and H.C.F. of two natural numbers

The product of L.C.M. and H.C.F. of two natural numbers = the product of the numbers.

Note. In particular, if Two natural numbers are co-prime then their L.C.M. = The product of the numbers.

HCF AND LCM of Fractions

1. HCF of Fractions = (HCF of Numerators/LCM of Denominators)

2. LCM of Fractions = (LCM of Numerators/HCF of Denominators)

Examples
1. Find the G.C.D of 12x2y3z2, 18x3y2z4, and 24xy4z3

(1) 6xy2z2
(2) 6x3y4z3
(3) 24xy2z2
(4) 18x2y2z3

Correct Choice is (1) and Correct Answer is 6xy2z2
________________________________________
Explanatory Answer

G.C.D of 12, 18 and 24 is 6.

The common factors are x, y, z and their highest powers common to all are 1, 2 and 2 respectively.

Therefore, G.C.D = 6xy2z2

2.Find the L.C.M. of 72, 240, 196.

Solution

Using Prime factorisation method

72 = 2×2×2×3×3 = 2³×3²
240 = 2×2×2×2×3×5 = 24×3×5
196 = 2×2×7×7 = 2²×7²

L.C.M. of the given numbers = product of all the prime factors of each of the given number with greatest index of common prime factors
= 24×3²×5×7² = 16×9×5×49 = 35280.

Using Division method

2 | 72, 240, 196
2 | 36, 120, 98
2 | 18, 60 , 49
3 | 9 , 30 , 49
| 3 , 10 , 49

L.C.M. of the given numbers
= product of divisors and the remaining numbers
= 2×2×2×3×3×10×49
= 72×10×49 = 35280.

3.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

4. Arrange the fractions 2/15, 3/10 , 5/21 in ascending order of their respective magnitudes.

Solution

5 | 15, 10, 21
3 | 3, 2, 21
| 1, 2 , 7

LCM of 15,10,21 = 5 x 3 x 2 x 7 = 210
Thus,

2/15 = (2x14)/(15x14) = 28/210

3/10 = (3x21)/(10x21) = 63/210

5/21 = (5x10)/(21x10) = 50/210

Thus now comparing these fractions is simple. Greater the Numerator, greater is the Fraction.
Thus ,

28/ 210 ; 50/210 ; 63/210 in ascending order

Or

2/15 ; 5/21 ; 3/10 in ascending order.


5. What is the smallest no which when increased by 3 is divisible by 27,35,25 and 21?
1] 4722
2] 4725
3] 4728
4] 4731

Solution:
The smallest no that is divisible by 27, 35, 25 and 21 = LCM of these nos.

So our answer is LCM – 3.

The LCM can be obtained by any of the techniques described above.

LCM = 4725

Ans 4722 ie Option 1.

6. What is the smallest no which when decreased by 5 is divisible by 36,48,21 and 28?
1] 1008
2] 1003
3] 1013
4] 1018

Solution

Same as Q 5.
In this case ans = LCM + 5
Ans option 3 – 1013

7. If A381 is divisible by 11, find the value of A?
1] 5 2] 6 3] 7 4] 8

Solution
As per the divisibility test of 11, we have A+8 – 3 – 1 should be divisible by 11.

So A + 4 should be a multiple of 11.
Thus A+4=11k where k is an integer.
Put k=0, A= -4 Not Possible.
k=-1, A= -15 Not Possible.
k=1 , A=7 ans . Option[3]
No other value of k is feasible.

8. A no ‘A’ is not divisible by 3. which of the following is definitely divisible by 3.
1] A^2 +1 2] A^2 – A 3] A^2 – 1 4] A^2 + A

Solution:
Consider, Option [3], (A+1)(A-1). Now as A is not divisible by 3 one of these (A+1) or (A-1) will be divisible by 3.
Thus Ans Option[3].

9. What is the greatest no which when divided by 6,7,8,9,10 leaves remainders as 4,5,6,7,8 respectively.

1] 997920 2] 997918 3] 999999 4] 997922

Solution
Actually for CAT u can directly mark option 2 by seeing that on dividing the no by 10 the remainder is 8. By divisibility test we know only, option 2 will leave a remainder 8 with 10.

Proper solution for this is :
6-4 = 2 7-5 = 2 8-6 = 2 9-7 = 2 10-8 = 2

LCM of 6,7,8,9,10 =2520
Greatest no of six digits = 999999
Greatest six digit no that is a multiple of 2520 = 2520 x 396 = 997920
Subtract 2 from this no to get the required answer – 997918 and which will give remainders as 4,5,6,7,8 when divided by 6,7,8,9,10.

Hence [2].

10. The HCF and LCM of two nos is given. It is possible to find out the two nos uniquely if
I. Either the sum or the difference between the two nos is known.
II. HCF of the two nos = LCM of the two Nos.
III. (LCM/HCF) = prime no.

1] I and II only.
2] II only
3] II and III only
4] I, II and III

Solution:

I We know HCF X LCM = Product of the nos = AB(Lets say)
Given that A-B or A+B is known.
Thus we have 2 equations and two variables, thus the soln can be uniquely determined.

II When HCF = LCM the two nos are equal. So again can be solved.

III LCM/HCF = prime no is known.

Then one of the nos = LCM and the other = HCF.
Thus III is also true

So As I , II , III all are true, ANS is option [4].

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