So, i have elaborated the concept of LCM and HCF and also written down the questions that are usually asked in these types of exams.

#### Concept of LCM and HCF:

1. **LCM (Least Common Multiple)**: The least number which is exactly divisible by each one of the number given. * For example*: LCM of 2,4,5 is 20.

2. **HCF (Highest Common Factor)** or **GCD (Greatest Common Divisor)**: It is the greatest number that divides the each of the number given. __ For example:__ HCF of 2,4,8 is 2 or HCF of 2,4,5 is 1 (

*here, 1 is the only number that divides all the given number i.e. 2,4,5*)

*Note:* If a number, let A divides the other number, let B exactly, i.e. B/A that means

*A is a factor of B*and

*B is a multiple of A*.

3. __ Co-primes:__ The two numbers are said to co-primes when there HCF is 1.

4. The product of 2 numbers = the product of LCM and HCF

– **N1 x N2 = LCM x HCF**

5. HCF of fractions (N1/N2) = **HCF of Numerator/LCM of denominator**

– LCM of fractions = **LCM of Numerator****/****HCF of denominator**

#### Questions on LCM and HCF for SSC and Bank exams:

For larger numbers- The HCF of 2923 and 3239

– Using the Formula of __HCF of Fractions__ = HCF of numerator/LCM of denominator

=> HCF of 9, 12, 18 and 21/ LCM of 10, 25, 35, 40 = **3/2800**

– Using the formula of __LCM of Fractions__ = LCM of numerator/HCF of denominator** => **LCM of 9, 12, 18, 21/ HCF of 10, 25, 35, 40 = **252/5 **

– We can write the numbers in fractions i.e. 175/10, 7/1 and 56/10

Now, using HCF and LCM of Fractions => HCF of fractions = **7/10** and LCM of fractions = **1400/1**

– Using formula = HCFxLCM = N1xN2 => 18×108 = 36xN2 => **N2 = 54** ^{4} x 3^{5} x 5^{2} x 7^{2}. Find the third number?

– 3240 = 2^{3} x 3^{4} x 5; 3600 = 2^{4} x 3^{2} x 52; HCF = 36 = 2^{2} x 3^{2}

Since, HCF is the product of lowest common factors, so third number must have (**2 ^{2} x 3^{2}**) as its factor.

Since, LCM is the product of highest common factors, so third number must have (

**3**) as its factor

^{5}x 7^{2}So, third number =

**2**

^{2}x 3^{5}x 7^{2}.

– Here, we have to find the HCF of 7m, 560cm and 910cm.

HCF of 700cm, 560cm and 910cm = **70**

– Here, we have to find the HCF so that least number of pens and pencils can be calculated which can be equally distributed among the students. Hence, the HCF of 1001 and 910 = **91**

– The LCM of 4,6,10 and 15 = 60

and the number will be **60+2 ( 2 is the remainder) = 62**