REAL NUMBERS (CLASS X - CBSE) Pre - exercise 1.2 examples

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Theorem 1.2 : (The Fundamental Theorem of Arithmetic)

Every composite number can be expressed  (factorized) as a product of  primes and this factorization is unique, apart from the order in which the prime factors occur.

Example - 5 : Consider the number 4ⁿ, where n is a natural number. Check whether there are any value of n for which 4ⁿ ends with the digit zero.

Solution : If the number 4ⁿ, for any n, were to end with the digit zero, then it would be divisible by 5 i.e., the prime factorization of 4ⁿ would contain the prime 5. This is not possible because,

                   4ⁿ = (2 x 2) ⁿ

So, the only prime in the factorization of 4ⁿ is 2.

So, the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 4ⁿ. So, there is no natural number n for which 4ⁿ ends with the digit zero.

Example - 6 : Find the LCM and HCF of 6 and 20 by the prime factorization method.

Solution :  

          Prime factors are:

          6  =  2  x  3           and   

          20  =  2  x  2  x  5  =  2²  x  5

HCF (6, 20)  =  2  [ Product of smallest power of each common prime factor in the numbers]

LCM (6, 20) = 2²  x  3  x  5  = 60 [ Product of the greatest power of each prime factor involved in the numbers]

From above,

          HCF (6, 20)   x   LCM (6, 20)  = 6  x  20

i.e., for any two positive integers a and b,

          HCF (a, b)   x   LCM (a, b)  = a  x  b

Example - 7 : Find HCF of 96 and 404 by the prime factorization method. Hence, find their LCM.

Solution :  The prime factorization of 96 and 404 gives,

          96  = 2⁵  x  3

          404  =  2²  x  101

\ HCF ( 96, 404)  =   2²  =  4

Also, LCM ( 96, 404)  =     96  x  404        =    96   x   404     =   9696

                                      HCF(96, 404)                4

Example - 8 : Find the HCF and LCM of 6, 72 and 120 using prime factorization method.

Solution :   Prime factors are:

            6    =  2 x 3

           72   =  2³  x  3² and

          120  =  2³  x  3  x  5

HCF (6, 72, 120)   =  2  x  3  =  6 and

LCM (6, 72, 120)  =  2³  x  3²  x  5  = 360

[NOTE: Product of three numbers is not equal to the product of their HCF and LCM]


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